Distance between tennis ball and the top of the net

[goaliejoe35]

Homework Statement

During a tennis match, a player serves the ball at 25.3 m/s, with the center of the ball leaving the racquet horizontally 2.50 m above the court surface. The net is 12.0 m away and 0.900 m high. When the ball reaches the net, (a) what is the distance between the center of the ball and the top of the net? (b) Suppose that, instead, the ball is served as before but now it leaves the racquet at 5.00° below the horizontal. When the ball reaches the net, what now is the distance between the center of the ball and the top of the net? Enter a positive number if the ball clears the net. If the ball does not clear the net, your answer should be a negative number.

I need help trying to set up an equation for this. I am not sure what I should use. could some one please point me in the right direction?

 

[konthelion]

You need to set up two equations, one for the distance in the x direction, and one for the distance in the y direction.
Use the equation: $$x = x_0 + v_0 t + (1/2) a t^2$$ (this can be applied in the y direction as well)

 

[goaliejoe35]

OK so would, x0= 0 and v0= 25.3 m/s? Also would t= 0.474s from the equation t=x/v = t=12/25.3 = 0.474s?

 

[goaliejoe35]

Wait I just confused myself even more. Is x equal to 12m? so plug that in and solve for t?

 

[konthelion]

Ok, now set up the equation for the y-direction. Hint: What is the initial velocity in the y? Also what is the initial height in y? What is the acceleration in this case?

Yeah, you want to find t because that's the time that it takes for it to reach 12m in the x-direction, then you can use that time and plug it back in the y-direction equation to find the height at time t. Then take the difference of that height with the height of the net, .9m

 

[goaliejoe35]

Would the initial velocity in the y be 0?

 

[konthelion]

Yes. Then the initial $$y_{0}$$ would be the height in which the player served the ball, which would be? Then write out the equation and plug back in t.

 

[goaliejoe35]

initial height would be 2.5 m

 

[goaliejoe35]

so wold the equation look like this?

y=2.5+0+.5(9.8)(.437)^2
y=3.44m

then subtract the height of the net from that to get 2.54m?

 

[konthelion]

Ok, then what would $$y = y_0 + v_0 t + (1/2) a t^2$$ be? a=acceleration due to gravity

Close, but no. In what direction is gravity in?

 

[goaliejoe35]

what I did above ^ ??

 

[goaliejoe35]

what do you mean? should it be negative?

 

[konthelion]

Yes, it should since you took the distance 2.5m to be positive, then the direction of gravity is downwards, negative. Then you should get y=? Then the difference between the ball and the net is?

 

[goaliejoe35]

y=2.5+0-.5(9.8)(.437)^2
y=1.56m

1.56-0.9= 0.66m above the net?

 

[konthelion]

Wait, no. Check your time,t, again. It's not correct. Look at your second post

 

[goaliejoe35]

The second post if wrong I think...

12=25.3t+4.9t^2 and solve for t and I get 0.437s

I don't think the way I did it in the second post is quite right.

 

[konthelion]

What? Then how are you getting t?

Dude, you're confusing yourself. Why are you plugging in 12? 12m is the distance in the x direction. You are finding Y. You were doing just fine, you're just mixing up the x and the y.

12=25.3t+4.9t^2

This is absolutely wrong. There are NO acceleration in the x direction. Reread the entire thread.

 

[goaliejoe35]

from the equation x=x0+v0t+.5gt^2 plug in x=12 v0=25.3 and solve for t

 

[goaliejoe35]

ohhhh my bad man i screwed up your right

 

[konthelion]

There are two equations
(1) $$y=2.5m-\frac{1}{2}(9.8m/s^2)(t)^2$$
(2) $$x=25.3m/s * t$$

You already solved for t, in (2) when x = 12. Now, plug it back into (1). Then, y would be?

 

[goaliejoe35]

y=2.5+0-.5(9.8)(.474)^2

y=1.4m

then 1.4-.9=.5m above the net

 

[konthelion]

Yes, that's correct.

 

[goaliejoe35]

Ok thanks for your help! For part B would i do basically the same ting just factor in the 5 degree angle?

 

[konthelion]

Yes. Do you know how to do that, you have to take the x, y-component forms of the velocity. For example, for the x-component form of the velocity = $$Vcos(\theta)$$, where theta is your angle.

 

[goaliejoe35]

and y is same thing but sin correct?

 

[konthelion]

Yep, that's correct.

 

[goaliejoe35]

Ok one other question after i get the components what do I do with them?

 

[goaliejoe35]

Find the magnitude?

 

[konthelion]

Just follow the same steps that you did in problem (a), except this time the velocities in x-y components is different (i.e. the y equation will now have an initial velocity whereas in part (a) it did not)

 

[goaliejoe35]

so do I use the same time?

 

[goaliejoe35]

No I don't, I use the x component in equation 1 and get the time then I plug the time back into equation 2 along with the y component and solve correct?

 

[goaliejoe35]

t=12/25.2 = .476s

y=2.5+2.21(.476)-.5(9.8)(.476)^2

y= 2.44

2.44-.9= 1.54m above the net correct?

 

[ace123]

Not exactly. I think you got your signs mixed up. It says 5 degrees below the horizontal. What would the components be. Write them out.

 

[ankurb]

You need to substitute -g for acceleration, since the 'direction' of g is downwards.