While standing on a bridge 15.0 m above the ground, you drop a stone from rest. when the stone has fallen 3.20 m, you throw a second stone straight down. what initial velocity must you give to the second stone if they are both to reach the ground at the same instant?

11.35 m/s

Explanation:

Let the first ball takes time t to reach the ground.

Use second equation of motion

Here, s = 15 m, a = 9.8 m/s^2, u = 0 m/s

By substituting the values

15 = 0 + 0.5 x 9.8 x t^2

t = 1.75 second

Now time taken by the first ball to reach 3.20 m is t'

Use second equation of motion

Here, s = 3.2 m, a = 9.8 m/s^2, u = 0 m/s

By substituting the values

3.2 = 0 + 0.5 x 9.8 x t'^2

t' = 0.81 s

Thus, the time for the second ball to reach the ground is t - t' = 1.75 - 0.81 = 0.94 s

For second ball:

Use second equation of motion

Here, s = 15 m, a = 9.8 m/s^2, u = ?, t = 0.94 s

By substituting the values

15 = u x 0.94 + 0.5 x 9.8 x 0.94 x 0.94

u = 11.35 m/s

Thus, the second ball has the initial velocity of 11.35 m/s.

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see earlier answer