Respuesta :
This is a projectile motionm with
Iniital height, Yo = 2.4 m
Iniital vertical velocity, Voy = 0
Kind of motion: vertical fall => Y = Yo - Voy * t - [g*t^2] / 2
Initial horizontal position, Xo =0
Initial horizontal velocity, Vox = 22.3 m/s
kind of motion: uniform in the horizontal direction => X = Vox *t
You can find:
a) Height when the ball is passing the net, which is at 11.9 m from the service point.
Time when X = 11.9 m
X = Vox*t => t = X / Vox = 11.9 m / 22.3m/s = 0.5336 s
Replace that time into the vertical equation to find the height, Y:
Y = Yo - Voy*t - g*(t^2) / 2 = 2.4m - 0 - 9.8 m/s^2 * (0.5336 s)^2 / 2 = 1.0 m
Then, given that height of the net is 0.91 m, the ball will pass 0.09m over the net (9 cm).
b) How far from the net the ball will hit the ground, given that it is 11.9 m away of the service point.
From the vertical equation you can determine the flight time:
Y = Yo - g(t^2) / 2 => 0 = 2.4 - 4.9 t^2 => 4.9t^2 = 2.4 = > t^2 = 2.4 / 4.9 = 0.49 => t = √0.49 = 0.7 s
Now use that time in the vertical equation to find the X position of the ball when it hits the ground:
X = Vox *t = 22.3m/s * 0.7 s = 15.6 m
Then, subtract the position of the net to find how far away from it the ball hits the ground: 15.6 - 11.9 =3.7 m.
Iniital height, Yo = 2.4 m
Iniital vertical velocity, Voy = 0
Kind of motion: vertical fall => Y = Yo - Voy * t - [g*t^2] / 2
Initial horizontal position, Xo =0
Initial horizontal velocity, Vox = 22.3 m/s
kind of motion: uniform in the horizontal direction => X = Vox *t
You can find:
a) Height when the ball is passing the net, which is at 11.9 m from the service point.
Time when X = 11.9 m
X = Vox*t => t = X / Vox = 11.9 m / 22.3m/s = 0.5336 s
Replace that time into the vertical equation to find the height, Y:
Y = Yo - Voy*t - g*(t^2) / 2 = 2.4m - 0 - 9.8 m/s^2 * (0.5336 s)^2 / 2 = 1.0 m
Then, given that height of the net is 0.91 m, the ball will pass 0.09m over the net (9 cm).
b) How far from the net the ball will hit the ground, given that it is 11.9 m away of the service point.
From the vertical equation you can determine the flight time:
Y = Yo - g(t^2) / 2 => 0 = 2.4 - 4.9 t^2 => 4.9t^2 = 2.4 = > t^2 = 2.4 / 4.9 = 0.49 => t = √0.49 = 0.7 s
Now use that time in the vertical equation to find the X position of the ball when it hits the ground:
X = Vox *t = 22.3m/s * 0.7 s = 15.6 m
Then, subtract the position of the net to find how far away from it the ball hits the ground: 15.6 - 11.9 =3.7 m.
(a) The ball passes the net at the height of [tex]\boxed{1.01\text{ m}}[/tex] and therefore, it does not touch the net.
(b) The ball will fall exactly at the distance of [tex]\boxed{15.58\;{\text{m}}}[/tex] from the point of hitting which is within the service court of [tex]18.3\text{ m}[/tex].
Further Explanation:
Let us first understand this situation properly; the tennis ball server hits the ball by the racket keeping it in the upright position and exactly at the time of contact, the ball were in air at the height of [tex]2.4\text{ m}[/tex] above the ground, as shown in the figure (a) attached below.
Now, after hitting , the ball leaves the contact and goes exactly in the horizontal direction with horizontal velocity [tex]{v_x}[/tex] of [tex]22.3\;{\text{m/s}}[/tex] and zero vertically velocity, goes straight.
As the ball moves ahead, it force of gravity pulls it down slowly making a curved path as shown in the figure (b) attached below.
Part (a):
In this case, at a horizontal distance of [tex]11.9\;{\text{m}}[/tex], there’s a net in front of the server, the ball is still in the air. For the ball to cross the net without touching it, the height of the ball after covering the horizontal distance of [tex]11.9\text{ m}[/tex] should be greater than the height of the net.
Time takebn by the ball to cover the distance of [tex]11.9\;{\text{m}}[/tex] is given by,
[tex]\begin{aligned}{\text{time}}&=\dfrac{{{\text{distance}}}}{{{\text{speed}}}}\\t&=\frac{{11.9\;{\text{m}}}}{{22.3\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}}\\&=0.53\;{\text{s}}\\\end{gathered}[/tex]
So, the ball takes [tex]0.53\text{ s}[/tex] to reach at the distance of [tex]11.9\text{ m}[/tex] with the speed of [tex]22.3\text{ m}/\text{s}[/tex].
Now, the vertical distance, the ball traverses in [tex]0.53\text{ s}[/tex], is given by
[tex]y = {u_y}t + \dfrac{1}{2}{a_y}{t^2}[/tex] …… (1)
Here, [tex]{u_y} = 0[/tex] (No vertical component at the time of contact)
Substitute [tex]9.8\text{ m}/\text{s}^2[/tex] for [tex]{a_y}[/tex] and [tex]0.53\,{\text{s}}[/tex] for [tex]t[/tex] in equation (1).
[tex]\begin{aligned}y&=(0)t+\dfrac{1}{2}(9.8\text{ m}/\text{s}^2)(0.53)^2\\&=1.39\text{ m}\end{aligned}[/tex]
Now, this means the ball is [tex]1.39\text{ m}[/tex] below the point of hitting.
Now by subtracting [tex]1.39\text{ m}[/tex] from the total height of [tex]2.4\text{ m}[/tex], we get the height [tex]h[/tex] of the ball from the ground when it reaches the net.
[tex]\begin{aligned}h&=(2.4 - 1.39){\text{m}}\\&= 1}}{\text{.01m}}\end{aligned}[/tex]
Therefore, the ball passes over the net because the net’s height is [tex]0.91\text{ m}[/tex].
Part (b):
Now to find whether the ball will land on the service court .i.e. within the distance of [tex]6.4\text{ m}[/tex] on the other side of the net, we first find the total horizontal Range of the ball.
Given:
The horizontal distance of the net from the point of hitting is [tex]11.9\text{ m}[/tex].
The service court is extended to [tex]6.4\text{ m}[/tex] from the position of net.
Concept:
The service court is exactly extended to the total horizontal distance of [tex](11.9+ 6.4)\text{ m}=18.3\text{ m}[/tex] from the point of hitting.
Now, the horizontal range is defined as the total horizontal distance covered by the projectile (ball in this case), and that is the product of the constant horizontal velocity and the time of flight.
[tex]R={v_x}\times T[/tex]
Here, T is the time of flight of the ball.
The vertical distance covered by the ball from the point of hitting to reach the ground is given by
[tex]y={u_y}T+\dfrac{1}{2}{a_y}{T^2}[/tex]
Substituting [tex]2.4\text{ m}[/tex] for [tex]y[/tex], [tex]9.8\text{ m/}\text{s}^2[/tex] for [tex]{a_y}[/tex] and [tex]0\text{ m/s}[/tex] for [tex]{u_y}[/tex] in above expression.
[tex]\begin{aligned}2.4&=(0)t+\dfrac{1}{2}(9.8){T^2}\\2.4&=\dfrac{1}{2}(9.8){T^2}\\{T^2}&=0.489\\T&=0.69\text{ s}\\\end{aligned}[/tex]
Substitute [tex]22.3\text{ m/s}[/tex] for [tex]{v_x}[/tex], [tex]0.699\text{ s}[/tex] for [tex]T[/tex] in the above equation of horizontal range,
[tex]\begin{aligned}R&=(22.3{\text{ m/s)(0}}{\text{.699 s)}}\\&=15.58\text{ m}\\\end{aligned}[/tex]
Hence, we now know that the ball will fall exactly at the distance of [tex]\boxed{15.58\text{ m}}[/tex] from the point of hitting which is within the service court of [tex]18.3\text{ m}[/tex].
Learn more:
1. During soccer practise, maya kicked a soccer ball at 37 degree https://brainly.com/question/11023695
2. Motion of two balls under gravity https://brainly.com/question/10934170
3. Conservation of energy on a frictionless surface https://brainly.com/question/3943029
Answer Details:
Grade: high school
Subject: physics
Chapter: kinematics
Keywords:
tennis, tennis ball, projectile motion, acceleration due to gravity, racket, service court, horizontal direction, horizontal range, 2.4 m, 22.3 m/s, 6.4 m, served,back line, court.
