Your opponent, in a desperate attempt to return your wicked serve, makes a high lob of the tennis ball that causes it to land out of bounds on the asphalt course behind you. It then bounces so that it just barely clears a 3.90 m wall that is 2.32 m from where the ball hit the asphalt and that is separating your court from the next. In the adjacent court play is suddenly interrupted by the ball leaving your court and bouncing 2.50 m up into the action. How far horizontally, in meters, does the ball travel in this second bounce? Hint: Use the trajectory of the first bounce to get the horizontal speed, and assume that it remains unchanged between bounces.
To determine the horizontal distance traveled by the ball in the second bounce, we need to calculate its horizontal speed.
Let's break down the problem into steps:
Step 1: Calculate the horizontal speed of the ball during the first bounce:
We can use the information given about the height of the wall (3.90 m) and the distance from where the ball hit the asphalt to the wall (2.32 m) to determine how long the ball takes to reach the wall.
Using the formula for free-fall motion:
h = (1/2) * g * t^2
Where:
h = height (3.90 m)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Rearranging the formula to solve for time:
t = sqrt(2h / g)
Substituting the given height:
t = sqrt(2(3.90) / 9.8)
t = sqrt(0.795918367)
The time for the first bounce is approximately 0.8925 seconds.
Now, we can calculate the horizontal speed using the formula:
s = d / t
Where:
s = speed
d = distance (2.32 m)
t = time (0.8925 s)
Substituting the given values:
s = 2.32 / 0.8925
s = 2.6025 m/s
So, the horizontal speed of the ball during the first bounce is approximately 2.6025 m/s.
Step 2: Calculate the horizontal distance covered by the ball during the second bounce:
Since the horizontal speed remains unchanged between bounces, we can use the previously calculated speed to determine the distance traveled in the second bounce.
The time for the second bounce is given by the information that the ball bounced 2.50 m up into the air, which implies it reached its maximum height and fell back down. From this, we can use the equation for vertical motion:
h = (1/2) * g * t^2
Where:
h = height (2.50 m)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Rearranging the formula to solve for time:
t = sqrt(2h / g)
Substituting the given height:
t = sqrt(2(2.50) / 9.8)
t = sqrt(0.510204082)
The time for the second bounce is approximately 0.7146 seconds.
Now, we can calculate the horizontal distance using the formula:
d = s * t
Where:
d = distance
s = speed (2.6025 m/s)
t = time (0.7146 s)
Substituting the given values:
d = 2.6025 * 0.7146
d = 1.8595 meters
Therefore, the ball travels approximately 1.8595 meters horizontally in the second bounce.