Question

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

• A. 49 m/s
• B. 55.5 m/s
• C. 26.1 m/s
• D. 9.8 m/s

Answer 1

Answer: (C)

26.1 m/s

Time that the first ball takes to fall through the height can be found by the equation of motion-

$s=ut+1/2a{t}^2$

⟹$122.5=\frac{1}{2}\times 9.81\times t^2$

⟹$t=5s$

Hence the second ball must take $5s-2s=3s$ to travel down this height.

From $s=ut+1/2a{t}^2$

⟹$122.5=u\left(3\right)+\frac{1}{2}\times 9.81\times 5^2$

⟹$u=26.1m/s$