Projectile Motion and Tennis

[ae4jm]

[SOLVED] Projectile Motion and Tennis

Homework Statement

A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20.0 m/s at and angle of 5 degrees above the horizontal. The horizontal distance to the net is 7.0 m, and the net is 1.0 m high. Does the ball clear the net? If so, by how much? If not, by how much does it miss?

Homework Equations

VFx=VIx
X=VIx(T)
VFy=VIy+Ay(T)
Y=VIy(T)+.5Ay(T^2)
VFy^2=VIy^2+2A(Y)

The Attempt at a Solution

I took:
Y=VIy(T)+.5Ay(T^2)
-1=1.74t-4.9t^2
0=1.74t-4.9t^2+1
t=.6629

I then did:
X=VIx(T)
X=19.92(T)
X=13.21m

The answer in the book is yes the ball cleared the net by 1.01m. If the distance to the net is 7m and I'm getting 13.21m, then my answer shows that the ball cleared the net by far more than 1.01m. Please, what am I doing wrong?

 

[kdv]

Homework Statement

A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20.0 m/s at and angle of 5 degrees above the horizontal. The horizontal distance to the net is 7.0 m, and the net is 1.0 m high. Does the ball clear the net? If so, by how much? If not, by how much does it miss?

Homework Equations

VFx=VIx
X=VIx(T)
VFy=VIy+Ay(T)
Y=VIy(T)+.5Ay(T^2)
VFy^2=VIy^2+2A(Y)

The Attempt at a Solution

I took:
Y=VIy(T)+.5Ay(T^2)
-1=1.74t-4.9t^2
0=1.74t-4.9t^2+1
t=.6629

I then did:
X=VIx(T)
X=19.92(T)
X=13.21m

The answer in the book is yes the ball cleared the net by 1.01m. If the distance to the net is 7m and I'm getting 13.21m, then my answer shows that the ball cleared the net by far more than 1.01m. Please, what am I doing wrong?

they want the vertical distance by which the ball clears the net. Find the value of y when the ball is above the net. That result minus one meter will give you by how much the ball cleared the net

 

[Dick]

13.21m, assuming you did everything else right, is a value of x. The height by which you clear the net is going to be a value for y. The value of x you computed corresponds to the time when the ball reaches the level of the net (-1). Which is good, since 13.21m>7m, then the ball cleared the net. Reverse what you just did. Use the x equation to find at what time the ball reached the net and substitute that back into the y equation to figure out the height at the net.

 

[ae4jm]

After you pointed it out I see that I made a mistake with my equations.
So,
I found X's time to be .3514s. I plugged this into the y equation to get y=-.0063m when the ball reaches the net. Then, since the ball began at 2m above the ground and the net is 1m above the ground, I took Yinitial and solved there. I took 2-(-.0063)=2.0063m and then I subtracted the distance of the 1m, which is 1.0063m, or 1.01m.

Now, most of the time I would be satisfied with getting the correct answere, but it seems that in Physics, sometimes, a problem can be worked incorrectly but still get the correct answer. If you have the time I would appreciate it if you checked out my reasoning--did I follow the correct steps to obtain the correct answer?

Thanks greatly!

 

[kdv]

After you pointed it out I see that I made a mistake with my equations.
So,
I found X's time to be .3514s. I plugged this into the y equation to get y=-.0063m when the ball reaches the net. Then, since the ball began at 2m above the ground and the net is 1m above the ground, I took Yinitial and solved there. I took 2-(-.0063)=2.0063m and then I subtracted the distance of the 1m, which is 1.0063m, or 1.01m.

Now, most of the time I would be satisfied with getting the correct answere, but it seems that in Physics, sometimes, a problem can be worked incorrectly but still get the correct answer. If you have the time I would appreciate it if you checked out my reasoning--did I follow the correct steps to obtain the correct answer?

Thanks greatly!

Your time is fine. Your y value is wrong. If your y value was correct, the ball would be at 2m - 0.0063 m above the net (NOT 2 -(-0.0063)).

I find that if you set y =0 at the initial oposition, the ball is +0.007469 meter at the position of the net so at 2.007 m above ground when aligned with the net, therefore at 1.01 above the net.

 

[Dick]

I plug that time into your y equation and I get +0.0063m. Which as you say rounds to 1.01m above the net. I'm not sure why you had to do the subtract a negative thing. Did you reverse the y equation or just subtract backwards?

 

[ae4jm]

In my y equation I was using negative velocity and positive acceleration. I left it equal to delta y. Should I have plugged delta y into the equation by using -1, then moving that to the right side of the equation would have given me a positive answer?

 

[Dick]

I just did y=1.74*t-4.9*t^2 and plugged in t=0.3514s. What did you do?

 

[ae4jm]

I apologize for just getting back to everyone. I just got off of work and I was able to work on this problem again before class this evening.

I started my problem differently, similar to Dick.
Yf=Yi+Vi(T)-0.5(A)(T)^2
Yf=2+1.74(.3514)-0.5(9.8)(.3514)^2
Yf=2+.6114-.6051
Yf=2.0063m
Since this is the distance of 2.0063m from the ground and the top of the net is 1m from the ground, I did 2.0063m-1m=1.0063m.
Then, 1.0063m rounds to 1.07m clearance above the net.

I think that maybe until I get better use to the equation I should use Yf=Yi+... instead of deltaY=Vi(T)..., this way I'm sure that I do the delta Y and not just solve for Y and expect to get the same results.

I believe this looks better than my previous solutions? Thanks!

 

[Dick]

Yeah, it looks a little better than the other one because you don't have to keep translating the y origin around. But it's basically the same. Looks good.

 

[alexadrienne]

how did you find Viy in this problem?

 

[Dick]

how did you find Viy in this problem?

You use trig. You have an initial speed and angle. Split the velocity into components.