Question

A ball is projected upward at time t = 0.00 s, from a point on a roof 40 m above the ground and experiences negligible air resistance. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 84.9 m/s. Consider all quantities as positive in the upward direction. The velocity of the ball when it is 16 m above the ground is closest to

Answer 1

Gravitational acceleration = g = -9.81 m/s²

Initial height of the ball = H₁ = 40 m

Final height of the ball = H₂ = 16 m

Displacement of the ball = H

H = H₂ - H₁

H = 16 - 40

H = -24 m

Initial velocity of the ball = V₁ = 84.9 m/s

Velocity of the ball when it is 16 m above the ground = V₂

Time taken by the ball to reach 16 m above the ground = T

H = V₁T + gT²/2

-24 = 84.9T + (-9.81)T²/2

4.905T² - 84.9T - 24 = 0

T = 17.587 sec or -0.278 sec

Time cannot be negative.

T = 17.587 sec

V₂ = V₁ + gT

V₂ = 84.9 + (-9.81)(17.587)

V₂ = -87.63 m/s

Negative sign indicates the velocity of the ball is downwards.

Velocity of the ball when it is 16 m above the ground = -87.63 m/s