You serve a tennis ball from a height of 1.8 m above the ground. The ball leaves your racket with a speed of 19.4 m/s at an angle of 7.28° above the horizontal. The horizontal distance from the court's baseline to the net is 11.83 m, and the net is 1.07 m high. Neglect spin imparted on the ball as well as air resistance effects. Does the ball clear the net?
If yes, by how much? If not, by how much did it miss?
Write the equations for ball height Z and horizontal location X vs time. Assume the ball is served perpendicular to the net, "down the middle" (Y=constant), not to a side line.
X = 19.4 cos 7.28 t = 19.24 t
When the ball reaches the net, X = 11.83 m, so t = 0.6149 s
Y = 1.80 + 19.4 sin 7.28 t - (9.80/2)t^2
= 1.80 + 2.468 t - 4.90 t^2
Solve for the Y height when t = 0.6149 s when the ball reaches the net), and compare it to the height of the net.
I get Y to exceed the net height when it reaches the net.
The equation for Y should have been for the ball height, Z.
To determine whether the ball clears the net, we need to calculate its vertical displacement at the net height.
First, let's calculate the total time of flight for the ball. We can use the following formula:
time = 2 * (vertical velocity) / (acceleration due to gravity)
In this case, the vertical velocity is the component of the initial velocity perpendicular to the ground, given by:
vertical velocity = initial velocity * sin(theta)
where theta is the launch angle (7.28°), and the acceleration due to gravity is approximately 9.8 m/s^2.
vertical velocity = 19.4 m/s * sin(7.28°)
vertical velocity = 19.4 m/s * 0.1269
vertical velocity = 2.46 m/s
time = 2 * 2.46 m/s / 9.8 m/s^2
time = 0.5 s
Next, we can find the vertical displacement of the ball during this time using the formula:
vertical displacement = initial velocity * sin(theta) * time - (1/2) * acceleration due to gravity * time^2
vertical displacement = 19.4 m/s * sin(7.28°) * 0.5 s - (1/2) * 9.8 m/s^2 * (0.5 s)^2
vertical displacement = 2.46 m/s * 0.5 s - 4.9 m/s^2 * 0.25 s^2
vertical displacement = 1.23 m - 1.225 m
vertical displacement = 0.005 m or 0.005 * 100 cm = 0.5 cm
The vertical displacement of the ball at the net height is approximately 0.5 cm. Since the net is 1.07 m high, the ball clears the net by approximately:
net clearance = 1.07 m - 0.005 m
net clearance = 1.065 m
Therefore, the ball clears the net by approximately 1.065 meters.
To determine if the ball clears the net, we need to find the maximum height the ball reaches and compare it to the height of the net.
To find the maximum height, we can first analyze the vertical motion of the ball.
1. Determine the initial vertical velocity component:
v_vertical = v_initial * sin(theta)
where v_initial is the initial speed of the ball (19.4 m/s) and theta is the launch angle (7.28°).
v_vertical = 19.4 m/s * sin(7.28°) = 2.301 m/s
2. Use the kinematic equation to calculate the time it takes for the ball to reach maximum height:
v_vertical = v_initial_vertical + a * t
Since the ball's vertical velocity decreases due to gravity, the acceleration (a) is -9.8 m/s² (negative because it acts in the opposite direction of the initial velocity).
0 m/s = 2.301 m/s + (-9.8 m/s²) * t
Solve for t:
2.301 m/s = 9.8 m/s² * t
t = 2.301 m/s / 9.8 m/s² ≈ 0.235 s
3. Calculate the maximum height using another kinematic equation:
h = v_initial_vertical * t + (1/2) * a * t²
h = 2.301 m/s * 0.235 s + (1/2) * (-9.8 m/s²) * (0.235 s)²
h ≈ 0.270 m (or 27 cm)
Now that we have determined the maximum height the ball reaches, we can compare it to the height of the net, which is 1.07 m.
Since the maximum height of the ball (0.270 m) is less than the height of the net (1.07 m), the ball does not clear the net.
To find by how much the ball misses the net, we subtract the maximum height from the height of the net:
1.07 m - 0.270 m = 0.8 m
Therefore, the ball misses the net by approximately 0.8 meters.