Question

a ball is thrown directly downward with an initial speed of 9 m/s from a height 32 m. What is the velocity of ball just before hitting the ground take g=9.8 m/s^2​

Answer 1

Explanation:

u1 = 9m/s

u2 = v (let)

h = 32 m

g = 9.8m/s

We know that , v^2 - u^2 = 2as

so , v^2 = 2x9.8x32 - 81

v^2 = 627.2 - 81 = 546.2

So , v = under root 546.2

v is nearly equall to 24m/s