# "A woman throws a ball at a vertical wall..." Proj. Motion

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1. Oct 14, 2015

### toboldlygo

So, I've already answered parts (b) and (c), but I'm struggling with (d). Thanks in advance for any help!

1. The problem statement, all variables and given/known data

"A woman throws a ball at a vertical wall d = 6.0 m away. The ball is h = 3 m above ground when it leaves the woman's hand with an initial velocity of 16 m/s at 45°. When the ball hits the wall, the horizontal component of its velocity is reversed; the vertical component remains unchanged. (Ignore any effects due to air resistance.)

(d) How long was the ball in the air after it left the wall?"

I've got the height when the ball hits the wall hw= 7.62 m, and the time it took for the ball hit to the wall t = 0.53 sec. I know these are right because they've been graded as such.

2. Relevant equations

I think the relevant equation is $$y = y_0 + v_0t+1/2 g t^2$$.

3. The attempt at a solution

I've set y= -7.62, y0=0, and 1/2 g = -4.905; solve for t using the quadratic formula. I'm treating it as if it's a different problem after the ball hits the wall, so it's independent of whatever happens before the ball hits the wall. I don't know if that's right, but it made sense to split the whole problem into two parts: before the ball hits the wall, and after the ball hits the wall.

Last edited: Oct 14, 2015
2. Oct 14, 2015

### paisiello2

I assume the question is the time it takes to hit the ground after leaving the wall.

I wouldn't say they are independent exactly but you are on the right track with your formula. I would double check your signs and figure out what v0 is.

3. Oct 14, 2015

### toboldlygo

voy would just be sin(45)*16, right? And I double checked the signs (gravity and y are negative according to the coordinate system I'm using) and did the calculation five different times and then also programmed my calculator to do it to make sure, and I keep getting an answer that isn't accepted.

4. Oct 14, 2015

### paisiello2

OK, I guess it might be easier for you to break the problem into two parts. Why don't we call the point where the ball hits the wall "1".

Can you put your formula together at point 1?

5. Oct 14, 2015

### toboldlygo

I'm not quite sure what you mean by that. Do you want to know which formula I used to get the time the ball took to hit the wall? Or do you want me to put in the known quantities into the y equation (which gives me the answer 7.62 m)?

6. Oct 14, 2015

### paisiello2

No, I want you to put together the formula with the right inputs after the ball hits the wall at point 1.

7. Oct 14, 2015

### toboldlygo

I just did the problem correctly! I failed to account for gravity when calculating for vy. I was using viy, but I had to use vfy for the "after the ball hits the wall" equation. Thanks for helping me out!

8. Oct 14, 2015

### azizlwl

How can the vertical component remained unchanged in a flight?.

9. Oct 14, 2015

### toboldlygo

It can't, which is what I realized after staring at the problem for an hour and doing it twenty times haha