Question

A ball is rolled off the edge of a swimming pool and strikes the water 2.0m away horizontally and 0.30s later. What is the initial speed of the ball? What is its speed when it strikes the water?

Answer 1

Horizontal distance covered by the ball = R = 0.5 m

Time period = t = 0.5 sec

Initial velocity of the ball = V₁

The ball rolls off the edge of the pool, so the ball will have zero initial velocity in the y-direction.

Initial velocity of the ball in horizontal direction = V_1x = V₁

Initial velocity of the ball in vertical direction = V_1y = 0 m/s

There is no horizontal force acting on the ball, therefore the horizontal speed of the ball will remain constant.

Final velocity of the ball in horizontal direction = V_2x = V₁

Final velocity of the ball in vertical direction = V_2y

Acceleration in the vertical direction = g = 9.81 m/s²

In the horizontal direction,

R = V₁t

2 = V₁(0.5)

V₁ = 4 m/s

V_2y = V_1y + gt

V_2y = 0 + (9.81)(0.5)

V_2y = 4.905 m/s

Final speed of the ball = V₂

V₂ = 6.33 m/s

Initial speed of the ball = 4 m/s

Speed of the ball when it strikes the water = 6.33 m/s

answered by: anonymous