First slide

Vertical projection from top of a tower

Question

A ball is dropped from a bridge 122.5 m above a river. After the ball has been falling for 2s, a second
ball is thrown straight down after it. What must be the initial velocity of the second ball, so that both
ball hit the water at the same time?

Moderate

A

$40 {ms}^{- 1}$
B

$55.5 {ms}^{- 1}$
C

$26.1 {ms}^{- 1}$
D

$9.6 {ms}^{- 1}$

Solution

Let the ball hit water in t second.
For first ball,    $s = u t + \frac{1}{2} a t^{2}$
$122.5 = 0 + \frac{1}{2} \times 9.8 \times t^{2} = 4.9 t^{2}$
$\Rightarrow$    $t = \sqrt{\frac{122.5}{4.9}} = \sqrt{25} = 5 s$
For second ball, $122.5 = u (5 - 2) + \frac{1}{2} \times 9.8 \times (5 - 2)^{2}$
= 3u + 44.1
$\Rightarrow$ 3u = 122.5 ~ 44.1
3u = 78.4 $\Rightarrow$ u = $26.1 {ms}^{- 1}$

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