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?
26.5
To solve this problem, we can use the principles of projectile motion.
a) To find the minimum speed required to clear the net, we can use the equation for the height of the projectile at a given distance:
h = h_0 + v_0^2 * sin^2(θ) / (2g)
Where:
h = height of the projectile
h_0 = initial height (2.50 m in this case)
v_0 = initial velocity
θ = launch angle (in this case, 0° since the ball is hit horizontally)
g = acceleration due to gravity (9.8 m/s^2)
Since the ball needs to clear the 0.90 m-high net, the height of the projectile should be greater than 0.90 m. Thus, we can rearrange the equation to solve for the minimum velocity:
v_0 = sqrt((h - h_0) * 2g / sin^2(θ))
Plugging in the values:
h = 0.90 m
h_0 = 2.50 m
θ = 0°
g = 9.8 m/s^2
v_0 = sqrt((0.90 - 2.50) * 2 * 9.8 / sin^2(0°))
b) To find where the ball will land if it just clears the net, we can use the equations for horizontal motion:
d = v_0 * t
Where:
d = horizontal distance traveled by the ball (distance to the net = 15.0 m)
t = time of flight
Since the ball is hit horizontally, the initial velocity in the horizontal direction is constant, and there is no horizontal acceleration. Therefore, we can solve for the time of flight:
t = d / v_0
Plugging in the values:
d = 15.0 m
v_0 (minimum velocity we calculated)
c) To determine if the ball will be "good" in the sense that it lands within 7.0 m of the net, we need to check if the horizontal distance traveled is less than or equal to 7.0 m.
If the horizontal distance is less than or equal to 7.0 m, then the ball is "good." Otherwise, it is not "good."
d) To find the time the ball is in the air, we can use the equation:
t_total = 2 * t
This is because the time to reach the peak height is equal to the time of descent.
Let's calculate the answers step-by-step.
a) Plugging in the values:
v_0 = sqrt((0.90 - 2.50) * 2 * 9.8 / sin^2(0°))
v_0 = sqrt(-1.60 * 2 * 9.8 / 0)
As sin(0°) = 0, the denominator is 0. Therefore, it is not possible to calculate the required minimum speed to clear the net.
b) We cannot determine where the ball will land if it just clears the net because we do not have the minimum speed required to clear the net.
c) We cannot determine if the ball will land within 7.0 m of the net because we do not have the required minimum speed.
d) We cannot calculate the time the ball is in the air because we do not have the required minimum speed.
To solve these questions, we will use the principles of projectile motion. Let's break down each part:
Part a) To find the minimum speed required for the ball to clear the net, we can use the concept of vertical motion. The ball should be at least 0.90m + 2.50m = 3.40m high when it reaches the net. At this point, the vertical displacement is equal to zero (since it reaches its maximum height). Using the equation:
Δy = V₀y * t + (1/2) * a * t²,
where Δy is the vertical displacement, V₀y is the initial vertical velocity, -9.8 m/s² is the acceleration due to gravity, and t is the time it takes to reach this height, we can solve for V₀y.
Since Δy = 0, the equation becomes:
0 = V₀y * t + (1/2) * (-9.8) * t².
Next, we solve for t:
0 = t * (V₀y - 4.9 * t).
This equation represents a quadratic equation, which can be solved for t. Once we have t, we can find the horizontal distance (range) using:
range = V₀x * t,
where V₀x is the initial horizontal velocity. We can assume that the initial vertical velocity (V₀y) is zero for a horizontal serve.
Part b) To find where the ball will land if it just clears the net, we can use the horizontal motion formula:
range = V₀x * t.
Since we know the range (15 m) and the time (t), we can solve for V₀x.
Part c) To determine if the ball will land within 7.0m of the net, we need to compare the range with the desired distance of 7.0m.
Part d) To find the time the ball will be in the air, we can use the value of t obtained in part a).
Now, let's calculate the values:
Part a):
We are given:
Vertical displacement, Δy = 3.40m
Initial vertical velocity, V₀y = 0 m/s
Acceleration due to gravity, a = -9.8 m/s²
Using the formula:
Δy = V₀y * t + (1/2) * a * t².
Rearranging the equation, we get:
3.40 = 0 * t + (1/2) * (-9.8) * t².
3.40 = -4.9 * t².
t² = 3.40 / -4.9.
Taking the square root on both sides, we find:
t = √(3.40 / -4.9).
t ≈ 0.737 s (using a positive value since we only need the time taken).
Now, for part b):
We have:
Range, R = 15.0m
t = 0.737 s (from part a).
Using the formula:
range = V₀x * t,
we can solve for V₀x:
15.0 = V₀x * 0.737.
V₀x = 15.0 / 0.737 ≈ 20.36 m/s.
So, the minimum initial speed required for the ball to clear the net is approximately 20.36 m/s horizontally.
For part c):
We have:
Range, R = 15.0m.
Since 15.0m > 7.0m, the ball will indeed land within 7.0m of the net.
For part d):
We have:
Time, t ≈ 0.737 s (from part a).
Therefore, the ball will be in the air for approximately 0.737 seconds.
Hi = 2.5
h = .9
u = constant = 15/t
v = -g t
.9 = 2.5 + 0*t -4.9 t^2
so
t^2 = 1.6/4.9 = .3265
t = .5714 second
u = 15/t = 26.25 m/s
onward by yourself