At serve, a tennis player aims to hit the ball horizontally.
part a)What minimum speed is required for the ball to clear the 0.90m--high net about 15.0m from the server if the ball is "launched" from a height of 2.50m ?
part b)here (relative to server) will the ball land if it just clears the net?
part c)Will it be "good" in the sense that it lands within 7.0m of the net? yes or no ?
part d) How long will it be in the air?
tsm
Part a) To find the minimum speed required for the ball to clear the 0.90m-high net, we can use the principles of projectile motion. The horizontal and vertical motions of the ball are independent of each other, so we can consider them separately.
First, let's analyze the vertical motion of the ball. We know that the ball is launched from a height of 2.50m and must clear the net, which is 0.90m high. The maximum height reached by the ball will be equal to its launch height plus the net height, so the maximum height is 2.50m + 0.90m = 3.40m.
We can use the kinematic equation to find the minimum vertical velocity (v) at the top of the ball's trajectory:
v^2 = u^2 + 2as
Where:
- v = final vertical velocity (0 m/s at the top of the trajectory)
- u = initial vertical velocity (unknown)
- a = acceleration due to gravity (-9.8 m/s^2, considering downward direction)
- s = displacement (maximum height reached, 3.40m)
0^2 = u^2 + 2(-9.8 m/s^2)(3.40m)
0 = u^2 - 66.04 m^2/s^2
Solving for u, we have:
u^2 = 66.04 m^2/s^2
u = √(66.04 m^2/s^2)
u ≈ 8.12 m/s
Therefore, the minimum vertical velocity (upward) component required for the ball to clear the net is approximately 8.12 m/s.
Now, let's analyze the horizontal motion of the ball. We want to find the minimum horizontal velocity required for the ball to travel a distance of 15.0m. Since there are no vertical forces acting on the ball during its flight, the horizontal velocity remains constant.
The formula to find the horizontal velocity (v_x) is:
v_x = d / t
Where:
- v_x = horizontal velocity (unknown)
- d = distance (15.0m)
- t = time of flight (unknown)
To find the time of flight, we can use the vertical motion of the ball. The time it takes for the ball to reach its maximum height (turning point) and fall back down to the net height is the same as the time it takes for the ball to travel the horizontal distance of 15.0m.
Using the formula:
s = ut + 0.5at^2
Where:
- s = displacement (maximum height reached, 3.40m)
- u = initial vertical velocity (8.12 m/s, upward direction)
- a = acceleration due to gravity (-9.8 m/s^2, considering downward direction)
- t = time of flight (unknown)
3.40m = 8.12 m/s * t + 0.5 * (-9.8 m/s^2) * t^2
0.5 * (-9.8 m/s^2) * t^2 + 8.12 m/s * t - 3.40m = 0
Solving this quadratic equation for t, we find two possible solutions:
t ≈ 0.563s (ignoring the negative value)
t ≈ 1.14s
Therefore, the time of flight for the ball is approximately 1.14s.
Now, we can use the time of flight to find the horizontal velocity:
v_x = d / t
v_x = 15.0m / 1.14s
v_x ≈ 13.16 m/s
So, the minimum horizontal velocity required for the ball to clear the net and travel a distance of 15.0m is approximately 13.16 m/s.
Part b) Since the horizontal velocity remains constant during the flight, the ball will land at the same horizontal position as its launch point. In other words, it will land at the server's position.
Part c) To determine if the ball will land within 7.0m of the net, we need to compare the horizontal distance traveled by the ball with the net distance.
Since the horizontal velocity is 13.16 m/s and the time of flight is 1.14s (found in part a), we can calculate the horizontal distance traveled by the ball:
d = v_x * t
d = 13.16 m/s * 1.14s
d ≈ 14.99 m
The calculated horizontal distance is approximately 14.99m, which is less than 7.0m. Therefore, the ball will land within 7.0m of the net.
Part d) The time of flight, t, was found to be approximately 1.14s in part a. Therefore, the ball will be in the air for approximately 1.14 seconds.