Question

A stone is dropped into a river from a bridge 41.7 m above the water. Another stone is thrown vertically down 1.80 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?

Answer 1

Answer:

31.75 m/s

Explanation:

h = 41.7 m

Let the initial velocity of the second stone is u

Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.

For first stone:

Use second equation of motion

$h=ut+\frac{1}{2}gt^2$

Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7

So, 41.7= 0 + 0.5 x 9.8 x t^2

41.7 = 4.9 t^2

t = 2.92 s ..... (1)

For second stone:

Use second equation of motion

$h=ut+\frac{1}{2}gt^2$

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity

$h=u\left ( t-1.8 \right )+4.9\left ( t-1.8 \right )^2$    .... (2)

By equation the equation (1) and (2), we get

$41.7=1.12 u +4.9 \times 1.12^{2}$

u = 31.75 m/s