A stone is dropped into a river from a bridge 46.5 m above the water. Another stone is thrown vertically down 1.28 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone ?

1 Answer
Mar 3, 2018

Answer:

#17 ms^-1#

Explanation:

Suppose, the #2# nd stone was thrown with a velocity of #u# downwards,if it took time #t# to reach the height of the bridge,then we can write,

#46.5 =ut + 1/2 g t^2# ....1 (using, #s=ut+1/2g t^2#)

now,if the 1st stone after dropping(i.e it had no initial velocity) took time #t'# to reach the height of the bridge then,

#46.5 = 1/2 g t'^2#

Given, #t'-t=1.28#

So,comparing the above three equations we can write,

#ut + 1/2 g t ^2 = 1/2 g (t+1.28)^2#.....2

now,solving equation 1 and 2 we get, #u=17 ms^-1#