Question

A ball is rolling horizontally at 3.00 meters per second as it leaves the edge of a tabletop 0.750 meter above the floor. The ball lands on the floor 0.391 second after leaving the tabletop. What is the magnitude of the ball’s acceleration 0.200 second after it leaves the tabletop?

Answer 1

Answer:

The magnitude of the acceleration of the ball is 9.81m / ^ 2 down, as you can see this value corresponds to gravity.

Explanation:

This movement is known as semi-parabolic, consider the following to solve

1) horizontally (X) there is a constant movement, with zero acceleration.

2) vertically (Y) there is a constant acceleration normally of the gravity value.

3) from the moment the ball begins to fall until it touches the ground the gravity is the same, therefore the acceleration is the same.

4) the initial vertical speed is 0.

When performing a mathematical demonstration, it is found that the equation that define this motion as the follows

$Y=VoT+0.5at^{2}$

where

Y=  table height=0.75m

Vo=0=initial speed

t=time=0.391s

a=acceleration

Solving

$Y=0.5at^2\\a=\frac{Y}{(0.5)(t^2)} \\a=\frac{0.75m}{(0.5)((0.391)^2)} =9.81m/s^2$

The magnitude of the acceleration of the ball is 9.81m / ^ 2 down, as you can see this value corresponds to gravity.