A man drops a ball downwards from the roof of a tower of height $400 ext{ m}$. At the same time, another bal

Question

A man drops a ball downwards from the roof of a tower of height $400 	ext{ m}$. At the same time, another ball is thrown upwards with a velocity $50 	ext{ m/s}$ from the surface of the tower. They will meet at which height from the surface of the tower?

Accepted Answer

1. **Analyze Relative Motion:** Both balls are under the influence of the same acceleration due to gravity ($g$) acting downwards. Therefore, the relative acceleration between them is zero ($a_{rel} = g - g = 0$).
2. **Calculate Time to Meet ($t$):** The initial relative velocity is the sum of their speeds (since they move towards each other): $u_{rel} = 50 + 0 = 50 	ext{ m/s}$. The relative distance to cover is the tower height $H = 400 	ext{ m}$.
   $$ t = \frac{	ext{Relative Distance}}{	ext{Relative Velocity}} = \frac{400}{50} = 8 	ext{ s} $$
   .
3. **Calculate Height from Surface ($h$):** Use the second equation of motion for the ball thrown upwards ($u = 50 	ext{ m/s}$, $a = -g = -10 	ext{ m/s}^2$).
   $$ h = ut - \frac{1}{2}gt^2 $$
   $$ h = 50(8) - \frac{1}{2}(10)(8)^2 $$
   $$ h = 400 - 5(64) = 400 - 320 = 80 	ext{ m} $$
   Alternatively, calculate the distance fallen by the top ball ($h' = \frac{1}{2}gt^2 = 320 	ext{ m}$) and subtract from total height: $400 - 320 = 80 	ext{ m}$ .

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