A body is thrown vertically upward. What initial speed is required so that the distances traversed during the 5th second and 6th second of motion are equal?

Projectile Motion: Equal Distance in Consecutive Seconds

🚀 Projectile Motion Problem

A body is thrown vertically upward. What initial speed is required so that the distances traversed during the 5th second and 6th second of motion are equal?

🔍 Detailed Solution

Key Concept: Time symmetry at peak position

Step 1: Understand the condition
For distances in 5th and 6th seconds to be equal, the body must reach its peak exactly at the end of the 5th second.

\( t_{\text{peak}} = 5 \, \text{seconds} \)

Step 2: Relate peak time to initial velocity
Time to reach maximum height is given by:

\( t_{\text{peak}} = \frac{u}{g} \)

Step 3: Solve for initial velocity (u)
Given \( t_{\text{peak}} = 5 \, \text{s} \) and \( g = 9.8 \, \text{m/s}^2 \):

\( 5 = \frac{u}{9.8} \)
\( u = 5 \times 9.8 = 49 \, \text{m/s} \)

Step 4: Verify distances

  • 5th second: Body rises with decreasing speed
    Distance = Average velocity × time
    \( = \frac{(u - 4g) + 0}{2} = \frac{49 - 39.2}{2} = 4.9 \, \text{m} \)
  • 6th second: Body falls from rest
    Distance = \( \frac{1}{2}g(1)^2 = 4.9 \, \text{m} \)

🎯 Final Answer: 49 m/s

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