Question

- contust, S, A Stome is dropped into water from bridge 45m above the water Surface another Stone is throw verticuly down ward 1 second after the first Stone is dropped Both Stone Strite the water at the Same time what was initial speed of the Second stone?​

Answer 1

Step 1: Calculate the time it takes for the first stone to reach the water surface using the equation h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.

45 = (1/2)(9.8)t^2

t^2 = 45 * 2 / 9.8

t^2 = 9.18

t ≈ 3.03 s

Step 2: Since the second stone is thrown 1 second after the first stone is dropped, the time it takes for the second stone to reach the water surface is t - 1.

t - 1 ≈ 3.03 - 1

t - 1 ≈ 2.03 s

Step 3: Use the equation v = u + gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time.

v = 0 (since the final velocity is zero when the stone reaches the water surface)

u = ?

g = 9.8 m/s^2

t = 2.03 s

0 = u + (9.8)(2.03)

u = -9.8(2.03)

u ≈ -19.894 m/s