🚀 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
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.
Step 2: Relate peak time to initial velocity
Time to reach maximum height is given by:
Step 3: Solve for initial velocity (u)
Given \( t_{\text{peak}} = 5 \, \text{s} \) and \( g = 9.8 \, \text{m/s}^2 \):
\( u = 5 \times 9.8 = 49 \, \text{m/s} \)
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
0 Comments