At serve, a tennis player aims to hit the ball horizontally. What minimum speed is required for the ball to clear the 0.91 m high net about L = 15.1 m from the server if the ball is "launched" from a height of H = 2.55 m?
How far from the end of the service box (which is a distance of S = 7.20 m from the net) will the ball land if it just clears the net? Use the positive direction if the ball is good (i.e. lands within the service box). How long will it be in the air?
26.31 m/s
18.68 m
1.95 s
To calculate the minimum speed required for the ball to clear the net, we can use the concept of projectile motion.
1. First, let's calculate the vertical distance the ball needs to travel to clear the net. The height of the net is 0.91 m, and the ball is launched from a height of 2.55 m above the ground. So, the vertical distance is:
Vertical distance (d) = 2.55 m - 0.91 m = 1.64 m
2. The formula for the vertical distance in projectile motion is:
d = (v^2 * sin^2(theta)) / (2 * g)
where v is the initial velocity, theta is the launch angle, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the ball is aimed to hit horizontally, the launch angle (theta) is 0 degrees, and sin(0) is equal to 0. Therefore, the equation simplifies to:
d = (v^2 * 0) / (2 * g)
This means the ball only needs to be launched with a horizontal velocity, so the vertical distance calculation is not necessary.
3. The horizontal distance the ball will travel is determined by the formula:
Horizontal distance (L) = v * t
where t is the time the ball is in the air.
To find the minimum speed, we need to find the maximum time required for the ball to land within the service box (landing distance is 7.20 m).
4. Using the formula for time (t) in projectile motion:
Vertical distance (d) = (v^2 * sin^2(theta)) / (2 * g)
1.64 m = (v^2 * sin^2(0)) / (2 * 9.8 m/s^2)
Since sin(0) is equal to 0, the equation simplifies to:
1.64 m = (v^2 * 0) / (2 * 9.8 m/s^2)
This equation tells us that the vertical distance does not depend on the velocity.
5. Now, let's find the maximum time (t) by using the horizontal distance:
Horizontal distance (L) = v * t
15.1 m = v * t
t = 15.1 m / v
6. To find the time it takes for the ball to reach the landing distance (S = 7.20 m):
t = 7.20 m / v
7. Equating both equations for t:
15.1 m / v = 7.20 m / v
15.1 m = 7.20 m
This equation implies that v can be any value, as long as both sides of the equation are equal.
8. The minimum speed required will be 0 m/s or any positive value.
9. Since the ball is aimed to hit horizontally, it will land at a distance equal to the initial horizontal distance (L), which is 15.1 m.
10. The total time the ball is in the air can be calculated by substituting the value of L back into the horizontal distance equation:
Horizontal distance (L) = v * t
15.1 m = v * t
t = 15.1 m / v
Therefore, the time the ball will be in the air will be 15.1 m / v.
To find the minimum speed required for the ball to clear the net, we can use the principle of conservation of mechanical energy. At the top of the ball's trajectory, all of its potential energy is converted into kinetic energy (horizontal velocity). At this point, the potential energy is given by mass times gravitational acceleration times the height, mgh, and the kinetic energy is given by half the mass times the square of the velocity, (1/2)mv^2.
The potential energy at the top of the trajectory is mgh and the kinetic energy is (1/2)mv^2. Since energy is conserved, we can set these two equal to each other:
mgh = (1/2)mv^2
The mass cancels out, and we can solve for the velocity (v):
gh = (1/2)v^2
v^2 = 2gh
v = sqrt(2gh)
Given:
Height of the net (h) = 0.91 m
Height of the ball launch (H) = 2.55 m
Acceleration due to gravity (g) = 9.8 m/s^2
Plugging in the values:
v = sqrt(2 * 9.8 * (2.55 - 0.91))
v = sqrt(2 * 9.8 * 1.64)
v = sqrt(32.096)
v ≈ 5.67 m/s
So, the minimum speed required for the ball to clear the net is approximately 5.67 m/s.
To find how far from the end of the service box the ball will land, we can use the range formula:
R = (v^2 sin(2θ)) / g
where R is the range, v is the initial velocity, θ is the launch angle (which is 0 for a horizontal serve), and g is the acceleration due to gravity.
Given:
Initial velocity (v) = 5.67 m/s
Acceleration due to gravity (g) = 9.8 m/s^2
Launch angle (θ) = 0 radians (horizontal)
Plugging in the values:
R = (5.67^2 sin(2 * 0)) / 9.8
R = (5.67^2 * 0) / 9.8
Since sin(0) is 0, the numerator becomes 0.
R = 0 / 9.8
R = 0 meters
Therefore, the ball will land at the net itself.
To find how long the ball will be in the air, we can use the formula:
T = (2 * v * sin(θ)) / g
where T is the time of flight, v is the initial velocity, θ is the launch angle (which is 0 for a horizontal serve), and g is the acceleration due to gravity.
Given:
Initial velocity (v) = 5.67 m/s
Acceleration due to gravity (g) = 9.8 m/s^2
Launch angle (θ) = 0 radians (horizontal)
Plugging in the values:
T = (2 * 5.67 * sin(0)) / 9.8
Since sin(0) is 0, the numerator becomes 0.
T = 0 / 9.8
T = 0 seconds
Therefore, the ball will be in the air for 0 seconds since it lands at the net itself.