A tennis ball is struck such that it leaves the racket horizontally with a speed of 53 m/s. The ball clears the net and hits the ground a horizontal distance 29 m from where it was hit. How long was the ball in the air and how high was the ball above the ground when it is hit?
time= distance/velocity
Now, knowing time in air,how far does a ball fall in that time?
distance=1/2 g time^2
To find the time the ball was in the air, we can use the horizontal distance the ball traveled and the horizontal velocity.
The horizontal distance traveled by the ball, denoted as d_x, is 29 m.
The horizontal velocity of the ball, denoted as v_x, is the initial velocity of 53 m/s.
We can use the formula: distance = velocity × time, where distance is the horizontal distance, velocity is the horizontal velocity, and time is the time the ball was in the air.
Therefore, we can rearrange the formula to solve for time:
time = distance / velocity.
Substituting in the given values:
time = 29 m / 53 m/s = 0.547 seconds.
So, the time the ball was in the air is approximately 0.547 seconds.
To find the height of the ball when it is hit, we can use the vertical motion of the ball.
Since the ball was struck horizontally, there is no initial vertical velocity. The only force acting on the ball is gravity. Therefore, the vertical motion of the ball is a free fall.
The vertical displacement can be determined by using the formula:
vertical displacement = (initial vertical velocity × time) + (0.5 × gravitational acceleration × time^2),
where the initial vertical velocity is zero, the gravitational acceleration is approximately 9.8 m/s^2, and time is the time the ball was in the air.
Therefore, the vertical displacement simplifies to:
vertical displacement = 0.5 × gravitational acceleration × time^2.
Substituting in the given values:
vertical displacement = 0.5 × 9.8 m/s^2 × (0.547 s)^2 = 1.452 m.
So, the ball was approximately 1.452 meters above the ground when it was hit.