Calculating the Speed Needed for the Tennis Ball to Land at 20.0 m Beyond the Fence
A student wants to launch a tennis ball horizontally just over a 6.0 m high fence and have it land at approximately 20.0 m beyond the fence. What speed would the tennis ball need to have to land at this approximate distance?
To solve this problem, we need to consider the vertical and horizontal components of the motion separately.
First, we'll calculate the time taken for vertical displacement. Since the initial velocity of the ball is 0, we can use the equation:
h = ut + (1/2)gt²
Where:
h = vertical displacement = 6.0 m
u = initial velocity = 0
g = acceleration due to gravity = 9.8 m/s²
t = time taken for vertical fall
Plugging in the values, we get:
6.0 = 0 + (0.5)(9.8)t²
t = 1.1065 s
Next, we'll calculate the horizontal velocity needed for the ball to land at 20.0 m beyond the fence. Since the time taken for vertical displacement is equal to the time taken for horizontal displacement, we can use the formula:
horizontal displacement = horizontal velocity x time
Plugging in the values, we get:
20.0 = horizontal velocity x 1.1065
horizontal velocity = 18 m/s (approx.)
Therefore, the tennis ball would need to have a horizontal velocity of approximately 18 m/s to land at the desired distance.
What speed would the tennis ball need to have in order to land at approximately 20.0 m beyond the 6.0 m high fence?
Explanation: Time taken for vertical displacement will be equal to the time taken for horizontal displacement. Let it be t for vertical fall, initial velocity u = 0, h = ut + (1/2)gt². We find that the horizontal velocity needed for the ball to land at 20.0 m beyond the fence is approximately 18 m/s.