A tennis player hits a ball 1.6m above the ground. The ball leaves the racket with a speed of 12m/s at an angle 8.0 above the horizontal. The net is a horizontal distance of 7.0m away from the player and is 1.0m high.
a) does the ball clear the net? Explain how you determined this. You must do a correct numerical calculation .
b)If the ball clears the top of the net, by what vertical distance does it clear? If not, by what vertical distance does is miss.
To determine whether the ball clears the net, we need to analyze its trajectory and compare it to the height of the net.
a) To calculate the motion of the ball, we can break down the initial velocity into its horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component is affected by gravity.
Given:
Initial velocity (v0) = 12 m/s
Launch angle (θ) = 8.0 degrees
Height of the net (h_net) = 1.0 m
Horizontal distance to the net (d_net) = 7.0 m
First, let's find the vertical component of the initial velocity (v0y) using trigonometry:
v0y = v0 * sin(θ)
v0y = 12 m/s * sin(8.0°)
v0y ≈ 1.656 m/s
Next, we can calculate the time taken by the ball to reach the net using the vertical motion equation:
Δy = v0y * t + (1/2) * a * t^2
Where Δy is the vertical displacement (1.6 m - 1.0 m = 0.6 m), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time of flight.
0.6 m = 1.656 m/s * t + (1/2) * (-9.8 m/s^2) * t^2
Simplifying the equation, we get a quadratic equation:
-4.9 t^2 + 1.656 t - 0.6 = 0
Solving this quadratic equation, we find two possible values for t:
t ≈ 0.114 s (ignoring negative value)
t ≈ 0.248 s
Since we want to find whether the ball clears the net, we need to calculate its vertical position at the horizontal distance of 7.0 m.
At 7.0 m horizontal distance, the time (t_net) taken by the ball to reach the net can be found using the horizontal motion equation:
d_net = v0x * t_net
Where v0x is the horizontal component of the initial velocity (v0x = v0 * cos(θ)).
7.0 m = 12 m/s * cos(8.0°) * t_net
Solving for t_net:
t_net ≈ 0.576 s
Now, we can calculate the vertical position of the ball when it reaches the net:
y_net = v0y * t_net + (1/2) * a * t_net^2
y_net = 1.656 m/s * 0.576 s + (1/2) * (-9.8 m/s^2) * (0.576 s)^2
y_net ≈ 0.935 m
b) To determine if the ball clears the net, we compare the calculated vertical position at the net (y_net ≈ 0.935 m) to the height of the net (1.0 m).
Since the calculated vertical position (0.935 m) is less than the height of the net (1.0 m), the ball does not clear the net.
If the ball cleared the top of the net, the vertical distance it would have cleared would be the difference between the height of the net and the calculated vertical position at the net:
Vertical distance cleared = 1.0 m - 0.935 m ≈ 0.065 m.