Projectile Motion MCQs
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1. A particle travels according to the equation \(x = at^3\), \(y = bt^3\). The equation of the trajectory is:
A. \(y = \frac{b}{2a}x^2\)
B. \(y = \frac{a}{2b}x^2\)
C. \(y = \frac{b}{a}x\)
D. \(y = \frac{a}{b}x^3\)
Explanation:
Given: \[ x = at^3, \quad y = bt^3 \] To find the trajectory, eliminate \(t\) between these two equations.
From the equation for \(x\): \[ t^3 = \frac{x}{a} \] Substitute \(t^3\) into the equation for \(y\): \[ y = b \left(\frac{x}{a}\right) \] Simplify: \[ y = \frac{b}{a}x \] Therefore, the trajectory is a straight line, and the correct answer is: \[ \text{C. } y = \frac{b}{a}x. \]
Given: \[ x = at^3, \quad y = bt^3 \] To find the trajectory, eliminate \(t\) between these two equations.
From the equation for \(x\): \[ t^3 = \frac{x}{a} \] Substitute \(t^3\) into the equation for \(y\): \[ y = b \left(\frac{x}{a}\right) \] Simplify: \[ y = \frac{b}{a}x \] Therefore, the trajectory is a straight line, and the correct answer is: \[ \text{C. } y = \frac{b}{a}x. \]
2. Speed at the maximum height of a projectile is half of its initial speed \(u\). Its range on the horizontal plane is:
A. \(\frac{u^2}{3g}\)
B. \(\frac{u^2 \sqrt{3}}{2g}\)
C. \(\frac{u^2}{\sqrt{3}g}\)
D. \(\frac{u^2}{2g}\)
Explanation:
At the maximum height, the vertical component of the velocity is zero, so the horizontal component remains: \[ v_x = u \cos \theta. \] Given that the speed at the maximum height is half of the initial speed: \[ u \cos \theta = \frac{u}{2}. \] Solve for \(\cos \theta\): \[ \cos \theta = \frac{1}{2}. \] This implies: \[ \theta = 60^\circ. \] The range \(R\) is given by: \[ R = \frac{u^2 \sin 2\theta}{g}. \] Substitute \(\theta = 60^\circ\): \[ R = \frac{u^2 \sin 120^\circ}{g} = \frac{u^2 (\sqrt{3}/2)}{g} = \frac{u^2 \sqrt{3}}{2g}. \] Therefore, the correct answer is: \[ \text{B. } \frac{u^2 \sqrt{3}}{2g}. \]
At the maximum height, the vertical component of the velocity is zero, so the horizontal component remains: \[ v_x = u \cos \theta. \] Given that the speed at the maximum height is half of the initial speed: \[ u \cos \theta = \frac{u}{2}. \] Solve for \(\cos \theta\): \[ \cos \theta = \frac{1}{2}. \] This implies: \[ \theta = 60^\circ. \] The range \(R\) is given by: \[ R = \frac{u^2 \sin 2\theta}{g}. \] Substitute \(\theta = 60^\circ\): \[ R = \frac{u^2 \sin 120^\circ}{g} = \frac{u^2 (\sqrt{3}/2)}{g} = \frac{u^2 \sqrt{3}}{2g}. \] Therefore, the correct answer is: \[ \text{B. } \frac{u^2 \sqrt{3}}{2g}. \]
3. A cricket ball is hit for a six leaving the bat at an angle of \(45^\circ\) to the horizontal with kinetic energy \(k\). At the top of the trajectory, the kinetic energy of the ball is:
A. \(0\)
B. \(k\)
C. \(\frac{k}{2}\)
D. \(\frac{k}{\sqrt{2}}\)
Explanation:
At the top of the trajectory, the vertical component of velocity is zero, and only the horizontal component contributes to the kinetic energy. The horizontal velocity is: \[ v_x = u \cos \theta. \] Given \(\theta = 45^\circ\): \[ v_x = \frac{u}{\sqrt{2}}. \] The kinetic energy at the top is: \[ K = \frac{1}{2} m (v_x)^2. \] Substitute \(v_x = \frac{u}{\sqrt{2}}\): \[ K = \frac{1}{2} m \left(\frac{u^2}{2}\right) = \frac{k}{{2}}. \] Therefore, the correct answer is: \[ \text{C. } \frac{k}{{2}}. \]
At the top of the trajectory, the vertical component of velocity is zero, and only the horizontal component contributes to the kinetic energy. The horizontal velocity is: \[ v_x = u \cos \theta. \] Given \(\theta = 45^\circ\): \[ v_x = \frac{u}{\sqrt{2}}. \] The kinetic energy at the top is: \[ K = \frac{1}{2} m (v_x)^2. \] Substitute \(v_x = \frac{u}{\sqrt{2}}\): \[ K = \frac{1}{2} m \left(\frac{u^2}{2}\right) = \frac{k}{{2}}. \] Therefore, the correct answer is: \[ \text{C. } \frac{k}{{2}}. \]
4. A particle is projected from a horizontal floor with speed \(10 \, \text{m/s}\) at an angle \(30^\circ\) with the floor. Which of the following is correct?
A. Velocity of particle will be perpendicular to initial direction two seconds after projection.
B. Minimum speed of particle will be \(5 \, \text{m/s}\).
C. Displacement of particle after half a second will be \(35/4 \, \text{m}\).
D. None of these
Explanation:
(A) The velocity of the particle will not be perpendicular to the initial direction as it is not possible in this scenario.
(B) Minimum speed during motion is: \[ u \cos \theta = 10 \cdot \cos 30^\circ = 10 \cdot \frac{\sqrt{3}}{2} \neq 5 \, \text{m/s}. \] (C) After \(t = 0.5 \, \text{s}\), the displacement does not match \(35/4 \, \text{m}\).
Therefore, the correct answer is: \[ \text{D. None of these.} \]
(A) The velocity of the particle will not be perpendicular to the initial direction as it is not possible in this scenario.
(B) Minimum speed during motion is: \[ u \cos \theta = 10 \cdot \cos 30^\circ = 10 \cdot \frac{\sqrt{3}}{2} \neq 5 \, \text{m/s}. \] (C) After \(t = 0.5 \, \text{s}\), the displacement does not match \(35/4 \, \text{m}\).
Therefore, the correct answer is: \[ \text{D. None of these.} \]
5. A body is projected with a speed \(u\) at an angle to the horizontal to achieve maximum range. At the highest point, the speed is:
A. \(0\)
B. \(\frac{u}{2}\)
C. \(u\)
D. \(\frac{u}{\sqrt{2}}\)
Explanation:
At the highest point, the vertical component of velocity is zero, and only the horizontal component remains. For maximum range: \[ \theta = 45^\circ. \] The horizontal velocity is: \[ v_x = u \cos \theta = u \cdot \frac{\sqrt{2}}{2}. \] Therefore: \[ v_x = \frac{u}{\sqrt{2}}. \] The correct answer is: \[ \text{D. } \frac{u}{\sqrt{2}}. \]
At the highest point, the vertical component of velocity is zero, and only the horizontal component remains. For maximum range: \[ \theta = 45^\circ. \] The horizontal velocity is: \[ v_x = u \cos \theta = u \cdot \frac{\sqrt{2}}{2}. \] Therefore: \[ v_x = \frac{u}{\sqrt{2}}. \] The correct answer is: \[ \text{D. } \frac{u}{\sqrt{2}}. \]
6. Ratio of the ranges of bullets fired from a gun (constant muzzle speed) at angles \(\theta\), \(2\theta\), and \(4\theta\) is found to be \(x : 2 : 2\). The value of \(x\) will be:
A. 1
B. 2
C. 3
D. None of these
Explanation:
For the ranges to be equal at \(2\theta\) and \(4\theta\), the sum of the angles must satisfy: \[ 2\theta + 4\theta = 90^\circ. \] Thus: \[ \theta = 15^\circ. \] The ratio of ranges is proportional to \(\sin(2\theta)\), so: \[ R_1 : R_2 : R_3 = \sin 30^\circ : \sin 60^\circ : \sin 120^\circ. \] Simplify: \[ R_1 : R_2 : R_3 = \frac{1}{2} : \frac{\sqrt{3}}{2} : \frac{\sqrt{3}}{2}. \] Therefore: \[ x = 3. \] The correct answer is: \[ \text{C. 3. } \]
For the ranges to be equal at \(2\theta\) and \(4\theta\), the sum of the angles must satisfy: \[ 2\theta + 4\theta = 90^\circ. \] Thus: \[ \theta = 15^\circ. \] The ratio of ranges is proportional to \(\sin(2\theta)\), so: \[ R_1 : R_2 : R_3 = \sin 30^\circ : \sin 60^\circ : \sin 120^\circ. \] Simplify: \[ R_1 : R_2 : R_3 = \frac{1}{2} : \frac{\sqrt{3}}{2} : \frac{\sqrt{3}}{2}. \] Therefore: \[ x = 3. \] The correct answer is: \[ \text{C. 3. } \]
7. A particle is projected with a speed \(10 \sqrt{2} \, \text{m/s}\) making an angle \(45^\circ\) with the horizontal. Neglecting air resistance, after 1 second of projection: (\(g = 10 \, \text{m/s}^2\)):
A. The height of the particle above the point of projection is \(5 \, \text{m}\).
B. The height of the particle above the point of projection is \(10 \, \text{m}\).
C. The horizontal distance of the particle from the point of projection is \(5 \, \text{m}\).
D. The horizontal distance of the particle from the point of projection is \(15 \, \text{m}\).
Explanation:
For the vertical displacement: \[ y = u_y t - \frac{1}{2} g t^2, \] where \(u_y = u \sin \theta = 10 \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 10 \, \text{m/s}\). Substituting \(t = 1 \, \text{s}\): \[ y = 10 \cdot 1 - \frac{1}{2} \cdot 10 \cdot (1)^2 = 10 - 5 = 5 \, \text{m}. \] Therefore, the height is \(5 \, \text{m}\).
For the horizontal displacement: \[ x = u_x t, \quad u_x = u \cos \theta = 10 \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 10 \, \text{m/s}. \] Substituting \(t = 1 \, \text{s}\): \[ x = 10 \cdot 1 = 10 \, \text{m}. \] However, only the vertical displacement matches \(5 \, \text{m}\), so the correct answer is: \[ \text{A. The height is \(5 \, \text{m}\).} \]
For the vertical displacement: \[ y = u_y t - \frac{1}{2} g t^2, \] where \(u_y = u \sin \theta = 10 \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 10 \, \text{m/s}\). Substituting \(t = 1 \, \text{s}\): \[ y = 10 \cdot 1 - \frac{1}{2} \cdot 10 \cdot (1)^2 = 10 - 5 = 5 \, \text{m}. \] Therefore, the height is \(5 \, \text{m}\).
For the horizontal displacement: \[ x = u_x t, \quad u_x = u \cos \theta = 10 \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 10 \, \text{m/s}. \] Substituting \(t = 1 \, \text{s}\): \[ x = 10 \cdot 1 = 10 \, \text{m}. \] However, only the vertical displacement matches \(5 \, \text{m}\), so the correct answer is: \[ \text{A. The height is \(5 \, \text{m}\).} \]
8. A particle has initial velocity \(\vec{v} = 3 \hat{i} + 4 \hat{j} \, \text{m/s}\) and a constant force \(\vec{F} = 4 \hat{i} - 3 \hat{j} \, \text{N}\) acts on it. The path of the particle is:
A. Straight line
B. Parabolic
C. Circular
D. Elliptical
Explanation:
The motion of the particle is determined by the net acceleration caused by the force: \[ \vec{a} = \frac{\vec{F}}{m}. \] Since the force has components in both \(x\) and \(y\), the particle experiences accelerations in both directions. This leads to a parabolic trajectory.
Therefore, the correct answer is: \[ \text{B. Parabolic.} \]
The motion of the particle is determined by the net acceleration caused by the force: \[ \vec{a} = \frac{\vec{F}}{m}. \] Since the force has components in both \(x\) and \(y\), the particle experiences accelerations in both directions. This leads to a parabolic trajectory.
Therefore, the correct answer is: \[ \text{B. Parabolic.} \]
9. A projectile is fired with a speed \(u\) at an angle \(\theta = 60^\circ\) from the horizontal. The range of the projectile is given by:
A. \(\frac{u^2}{g}\)
B. \(\frac{u^2 \sqrt{3}}{2g}\)
C. \(\frac{u^2}{g \sqrt{3}}\)
D. \(\frac{u^2}{2g}\)
Explanation:
The range of a projectile is given by: \[ R = \frac{u^2 \sin 2\theta}{g}. \] Substituting \(\theta = 60^\circ\): \[ R = \frac{u^2 \sin 120^\circ}{g}. \] Since \(\sin 120^\circ = \sin 60^\circ = \sqrt{3}/2\), we have: \[ R = \frac{u^2 (\sqrt{3}/2)}{g} = \frac{u^2 \sqrt{3}}{2g}. \] Therefore, the correct answer is: \[ \text{B. } \frac{u^2 \sqrt{3}}{2g}. \]
The range of a projectile is given by: \[ R = \frac{u^2 \sin 2\theta}{g}. \] Substituting \(\theta = 60^\circ\): \[ R = \frac{u^2 \sin 120^\circ}{g}. \] Since \(\sin 120^\circ = \sin 60^\circ = \sqrt{3}/2\), we have: \[ R = \frac{u^2 (\sqrt{3}/2)}{g} = \frac{u^2 \sqrt{3}}{2g}. \] Therefore, the correct answer is: \[ \text{B. } \frac{u^2 \sqrt{3}}{2g}. \]
10. A particle is projected with a speed \(u = 10 \, \text{m/s}\) at an angle of \(45^\circ\) with the horizontal. The maximum height reached by the particle is:
A. \(2.5 \, \text{m}\)
B. \(5 \, \text{m}\)
C. \(7.5 \, \text{m}\)
D. \(10 \, \text{m}\)
Explanation:
The maximum height of a projectile is given by: \[ H = \frac{(u \sin \theta)^2}{2g}. \] Substituting \(u = 10 \, \text{m/s}\), \(\theta = 45^\circ\), and \(g = 10 \, \text{m/s}^2\): \[ H = \frac{(10 \cdot \sin 45^\circ)^2}{2 \cdot 10}. \] Since \(\sin 45^\circ = \frac{1}{\sqrt{2}}\): \[ H = \frac{(10 \cdot \frac{1}{\sqrt{2}})^2}{20} = \frac{100 \cdot \frac{1}{2}}{20} = 2.5 \, \text{m}. \] Therefore, the correct answer is: \[ \text{A. } 2.5 \, \text{m}. \]
The maximum height of a projectile is given by: \[ H = \frac{(u \sin \theta)^2}{2g}. \] Substituting \(u = 10 \, \text{m/s}\), \(\theta = 45^\circ\), and \(g = 10 \, \text{m/s}^2\): \[ H = \frac{(10 \cdot \sin 45^\circ)^2}{2 \cdot 10}. \] Since \(\sin 45^\circ = \frac{1}{\sqrt{2}}\): \[ H = \frac{(10 \cdot \frac{1}{\sqrt{2}})^2}{20} = \frac{100 \cdot \frac{1}{2}}{20} = 2.5 \, \text{m}. \] Therefore, the correct answer is: \[ \text{A. } 2.5 \, \text{m}. \]
11. A ball is thrown upward at \(30^\circ\) with a speed of \(15 \, \text{m/s}\). What is the time of flight? (\(g = 10 \, \text{m/s}^2\)):
A. \(1 \, \text{s}\)
B. \(1.5 \, \text{s}\)
C. \(2 \, \text{s}\)
D. \(3 \, \text{s}\)
Explanation:
The time of flight is given by: \[ T = \frac{2u \sin \theta}{g}. \] Substituting \(u = 15 \, \text{m/s}\), \(\theta = 30^\circ\), and \(g = 10 \, \text{m/s}^2\): \[ T = \frac{2 \cdot 15 \cdot \sin 30^\circ}{10}. \] Since \(\sin 30^\circ = \frac{1}{2}\): \[ T = \frac{2 \cdot 15 \cdot \frac{1}{2}}{10} = \frac{15}{10} = 1.5 \, \text{s}. \] Therefore, the correct answer is: \[ \text{B. } 1.5 \, \text{s}. \]
The time of flight is given by: \[ T = \frac{2u \sin \theta}{g}. \] Substituting \(u = 15 \, \text{m/s}\), \(\theta = 30^\circ\), and \(g = 10 \, \text{m/s}^2\): \[ T = \frac{2 \cdot 15 \cdot \sin 30^\circ}{10}. \] Since \(\sin 30^\circ = \frac{1}{2}\): \[ T = \frac{2 \cdot 15 \cdot \frac{1}{2}}{10} = \frac{15}{10} = 1.5 \, \text{s}. \] Therefore, the correct answer is: \[ \text{B. } 1.5 \, \text{s}. \]
12. A projectile is launched with a speed \(20 \, \text{m/s}\) at an angle of \(60^\circ\). What is the horizontal range of the projectile? (\(g = 10 \, \text{m/s}^2\))
A. \(20 \, \text{m}\)
B. \(40 \, \text{m}\)
C. \(34.6 \, \text{m}\)
D. \(60 \, \text{m}\)
Explanation:
The horizontal range is given by: \[ R = \frac{u^2 \sin 2\theta}{g}. \] Substituting \(u = 20 \, \text{m/s}\), \(\theta = 60^\circ\), and \(g = 10 \, \text{m/s}^2\): \[ R = \frac{20^2 \sin(120^\circ)}{10}. \] Since \(\sin 120^\circ = \sin 60^\circ = \sqrt{3}/2\): \[ R = \frac{400 \cdot \sqrt{3}/2}{10} = 34.6 \, \text{m}. \] Therefore, the correct answer is: \[ \text{C. } 34.6 \, \text{m}. \]
The horizontal range is given by: \[ R = \frac{u^2 \sin 2\theta}{g}. \] Substituting \(u = 20 \, \text{m/s}\), \(\theta = 60^\circ\), and \(g = 10 \, \text{m/s}^2\): \[ R = \frac{20^2 \sin(120^\circ)}{10}. \] Since \(\sin 120^\circ = \sin 60^\circ = \sqrt{3}/2\): \[ R = \frac{400 \cdot \sqrt{3}/2}{10} = 34.6 \, \text{m}. \] Therefore, the correct answer is: \[ \text{C. } 34.6 \, \text{m}. \]
13. A ball is thrown vertically upward with a speed of \(20 \, \text{m/s}\). How long does it take to reach the highest point? (\(g = 10 \, \text{m/s}^2\))
A. \(1 \, \text{s}\)
B. \(2 \, \text{s}\)
C. \(3 \, \text{s}\)
D. \(4 \, \text{s}\)
Explanation:
At the highest point, the vertical velocity becomes zero. Using the first equation of motion: \[ v = u - g t. \] Substituting \(v = 0\), \(u = 20 \, \text{m/s}\), and \(g = 10 \, \text{m/s}^2\): \[ 0 = 20 - 10t \implies t = 2 \, \text{s}. \] Therefore, the correct answer is: \[ \text{B. } 2 \, \text{s}. \]
At the highest point, the vertical velocity becomes zero. Using the first equation of motion: \[ v = u - g t. \] Substituting \(v = 0\), \(u = 20 \, \text{m/s}\), and \(g = 10 \, \text{m/s}^2\): \[ 0 = 20 - 10t \implies t = 2 \, \text{s}. \] Therefore, the correct answer is: \[ \text{B. } 2 \, \text{s}. \]
14. A particle is projected horizontally from the top of a tower with a speed of \(10 \, \text{m/s}\). If the height of the tower is \(45 \, \text{m}\), find the time it takes to hit the ground. (\(g = 10 \, \text{m/s}^2\))
A. \(2 \, \text{s}\)
B. \(3 \, \text{s}\)
C. \(3.5 \, \text{s}\)
D. \(4 \, \text{s}\)
Explanation:
The time to hit the ground depends on the vertical motion. Using: \[ h = \frac{1}{2} g t^2. \] Substituting \(h = 45 \, \text{m}\) and \(g = 10 \, \text{m/s}^2\): \[ 45 = \frac{1}{2} \cdot 10 \cdot t^2 \implies 45 = 5t^2 \implies t^2 = 9 \implies t = 3 \, \text{s}. \] Therefore, the correct answer is: \[ \text{C. } 3.5 \, \text{s}. \]
The time to hit the ground depends on the vertical motion. Using: \[ h = \frac{1}{2} g t^2. \] Substituting \(h = 45 \, \text{m}\) and \(g = 10 \, \text{m/s}^2\): \[ 45 = \frac{1}{2} \cdot 10 \cdot t^2 \implies 45 = 5t^2 \implies t^2 = 9 \implies t = 3 \, \text{s}. \] Therefore, the correct answer is: \[ \text{C. } 3.5 \, \text{s}. \]
15. A ball is thrown horizontally from a height \(H\) with a speed \(v_0\). What is the horizontal range of the ball?
A. \(\sqrt{\frac{2H}{g}}\)
B. \(v_0 \sqrt{\frac{2H}{g}}\)
C. \(\frac{H}{g}\)
D. \(v_0^2 \cdot H\)
Explanation:
The horizontal range is given by: \[ R = v_0 \cdot t, \] where \(t\) is the time of flight. From vertical motion: \[ H = \frac{1}{2} g t^2 \implies t = \sqrt{\frac{2H}{g}}. \] Substitute \(t\) into the range equation: \[ R = v_0 \cdot \sqrt{\frac{2H}{g}}. \] Therefore, the correct answer is: \[ \text{B. } v_0 \sqrt{\frac{2H}{g}}. \]
The horizontal range is given by: \[ R = v_0 \cdot t, \] where \(t\) is the time of flight. From vertical motion: \[ H = \frac{1}{2} g t^2 \implies t = \sqrt{\frac{2H}{g}}. \] Substitute \(t\) into the range equation: \[ R = v_0 \cdot \sqrt{\frac{2H}{g}}. \] Therefore, the correct answer is: \[ \text{B. } v_0 \sqrt{\frac{2H}{g}}. \]
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