Question

A stone is dropped into water from a bridge 441m above the water. Another stone is thrown vertically 1 sec later. Both strike the water simultaneously. What was the initial speed of the second stone?

Answer 1

Let ‘t’ be the time taken by both the stones to reach the water.

For the stone that is dropped, using,

S = ut + ½ at²

=> 441 = 0 + ½ (9.8)t²

=> t = 9.5 s

Let the other stone be thrown at initial speed ‘u’. It reaches the water in (9.5 – 1 =) 8.5 s.

Using,

S = ut + ½ at²

=> 441 = 8.5u + ½ (9.8)(8.5²)

=> u = 10.2 m/s

So, the second stone is thrown at 10.2 m/s.