Question

in a tall building, a certain window is 500 meters above the ground.A tennis ball is catapulted horizontally out of the window with a horizontal initial velocity of 4 m/s. How far from the base of the building will the ball land?

Answer 1

The tennis ball lands at a point 40.4 m from the base of the building.

The tennis ball is projected with a horizontal velocity u from a window, which is at a height y from the ground. The ball lands at a distance x from the base of the building. Let the ball take a time t to reach the ground. In the time t ,the ball falls a vertical distance y and also travel a horizontal distance x.

The initial vertical velocity of the ball is zero, since the ball is projected in the horizontal direction.  The ball falls down under the action of gravitational force.

Thus, use the equation of motion,

$y=\frac{1}{2} gt^2$

rewrite the expression for t and calculate the value of t using 9.81 m/s²for g and 500 m for y.

$t=\sqrt{\frac{2y}{g} } \\ =\sqrt\frac{(2)(500m)}{9.81m/s^2} \\ =10.096 s$

The horizontal distance x is traveled using the constant velocity u since no force acts on the ball in the horizontal direction.

Therefore,

$x=ut$

Substitute 4 m/s for u and 10.096 s for t

$x=ut\\ =(4m/s)(10.096s)\\ =40.384m=40.4 m$

Thus, the ball lands at a point 40.4 m from the base of the building.