# Math Help - Motion Projectile, Physics

1. ## Motion Projectile, Physics

A tennis ball is struck @ the baseline of the court, 12m from the net. The ball is given an initial velocity with a horizontal component of 24 m/s an initial elevation of 1m

A) What vertical component of initial velocity must be given to the ball, such that it barely clears the 1m high net?

B) How far beyond the net will the ball hit the ground

The answers are 4.46 m/s and 10.9m respectively

2. Originally Posted by realintegerz
A tennis ball is struck @ the baseline of the court, 12m from the net. The ball is given an initial velocity with a horizontal component of 24 m/s an initial elevation of 1m

A) What vertical component of initial velocity must be given to the ball, such that it barely clears the 1m high net?

B) How far beyond the net will the ball hit the ground

The answers are 4.46 m/s and 10.9m respectively
A) From the horizontal motion it should be evident that the ball reaches the net when t = 0.5 s.

Motion in vertical direction:

a = -9.8 m/s^2
s = 0 m
t = 0.5 s
u = ?

Substitute the data into $s = ut + \frac{1}{2} at^2$ and solve for u.

B) Motion in vertical direction:

a = -9.8 m/s^2
s = -1 m
u = substitute answer from A)
t = ?

Substitute the data into $s = ut + \frac{1}{2} at^2$ and solve for t.

Motion in horizontal direction:

u = 24 m/s
a = 0 m/s^2
t = substitute from above.
s = ?

Substitute the data into $s = ut + \frac{1}{2} at^2$ and solve for s.

Then the answer is s - 12.

3. Originally Posted by realintegerz
A tennis ball is struck @ the baseline of the court, 12m from the net. The ball is given an initial velocity with a horizontal component of 24 m/s an initial elevation of 1m

A) What vertical component of initial velocity must be given to the ball, such that it barely clears the 1m high net?

B) How far beyond the net will the ball hit the ground

The answers are 4.46 m/s and 10.9m respectively
The time the ball would reach the net,
24t = 12
t = 0.5 sec

Just to clear the net,
y = 2.0 -1.0 = 1.0 meter more for the ball to climb vertically
Let V = vertical component of the initial velocity.
y = V*t -(1/2)g*t^2
1.0 = V(0.5) -4.9(0.5)^2
1.0 +1.225 = V(0.5)
V = 2.225 /0.5 = 4.45 m/sec

---------------------------
How far beyond the net will the ball hit the ground?

It should be more than 12.0 meters...because when the ball is 1,0 meter above the ground on the other side of the net, the ball should have traveled horizontally the same 12.0 meters it traveled approaching the net. That is in the parabolic property of the flight of the ball.

So, the answer 10.9 meters is wrong.

Basing from the 1,0 meter level where the ball is hit initially, the ball should fall down another 1.0 meter to hit the ground.
Thus, y = -1.0 meter
Hence, if t' = time for the ball to touch the ground after being hit,
-1.0 = (4.45)*t' -4.9(t')^2
4.9(t')^2 -4.45(t') -1.0 = 0
t' = {4.45 +,-sqrt[(-4.45)^2 -4(4.9)(-1.0)]} / 2(4.9)
t' = 1.09461 sec, or -0.18644 sec

Hence, the ball should travel a horizontal distance of
d = r*t
d = (24 m/sec)(1.09461sec) = 26.27 meters

Therefore, the ball would hit the ground at 26.27 -12.0 = 14.27 meters beyond the net.

4. Originally Posted by ticbol
The time the ball would reach the net,
24t = 12
t = 0.5 sec

Just to clear the net,
y = 2.0 -1.0 = 1.0 meter more for the ball to climb vertically
[snip]
Have I missed something here ....? The ball has an initial elevation of 1m (measured from the ground I assume). The net is 1m high. So I would think that the vertical displacement of the ball from where it's hit to the top of the net is 1 - 1 = 0 m .....?

5. Originally Posted by mr fantastic
Have I missed something here ....? The ball has an initial elevation of 1m (measured from the ground I assume). The net is 1m high. So I would think that the vertical displacement of the ball from where it's hit to the top of the net is 1 - 1 = 0 m .....?
Oppsss, the net is 1.0 m high only? I thought it was 2.0 m high all along.
Tennis! Oh, my.
Many a times I attack a problem without me thinking first if the given facts are practical.

So my computations are wrong.

Okay, let me check what would have happened.

The time the ball would reach the net,
24t = 12
t = 0.5 sec

Just to clear the net,
meaning the maximum height of the ball is midway of the 12-m distance of the hitter and the net. And the t there is 0.5 /2 = 0.25 sec.
At that point, the vertical velocity is zero, so,
Let V = vertical component of the initial velocity.
0 = V -gt
0 = V -9.8(0.25)
V = 2.45 m/sec ------**

---------------------------
How far beyond the net will the ball hit the ground?

Basing from the 1.0 meter level where the ball was hit initially, the ball should fall down another 1.0 meter to hit the ground.
Thus, y = -1.0 meter
Hence, if t' = time for the ball to touch the ground after being hit,
-1.0 = (2.45)t' -4.9(t')^2
4.9(t')^2 -2.45(t') -1.0 = 0
t' = {2.45 +,-sqrt[(-2.45)^2 -4(4.9)(-1.0)]} / 2(4.9)
t' = 0.76632 sec, or -0.0.26632 sec

Hence, the ball should travel a horizontal distance of
d = r*t
d = (24 m/sec)(0.76632 sec) = 18.39 meters

Therefore, the ball would hit the ground at 18.39 -12.0 = 6.39 meters beyond the net.

6. Is there any way you can get the answer to be 10.9 m for b???

cause i am lost and have a headache...how ironic (smiley)