# catch

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 23 m/s at an angle 33 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

After catching the ball, Sarah throws it back to Julie. The ball leaves Sarah's hand a distance 1.5 meters above the ground, and is moving with a speed of 23 m/s when it reaches a maximum height of 11 m above the ground. The speed of the ball when it leaves Sarah's hand is 26.75m/s.

How high above the ground will the ball be when it gets to Julie?
I think I have input everything value in the right place and get 1.5 meters; but it was not the correct answer, why?

 jcaron2 Posts: 983, Reputation: 1034 Senior Member #2 Jan 25, 2011, 10:17 PM
So how did you calculate it? If you tell me how you approached the problem, I might be able to tell you where you made a mistake.

For what it's worth, I did a quick calculation and I think the answer is about 8.24 m.
 western50 Posts: 105, Reputation: 12 New Member #3 Jan 25, 2011, 11:58 PM
so I use y=y,initial+v,initial*t+0.5at^2... y is what I want to get. y-initial is 1.5m, v-initial is 26.75m/s, t=4.68s( I got it by t(max)=velocity at the top/g), and because the ball moves in parabola, double the time). That's how I did it.

that's what I remembered
 jcaron2 Posts: 983, Reputation: 1034 Senior Member #4 Jan 26, 2011, 08:39 AM

Good. That equation will be used several times in the solution of this problem. However, I see a couple of issues with your approach. Remember that it's only the vertical component of the velocity that matters as far as the time the ball stays in the air, not the total velocity. Also, doubling the time because of the parabolic trajectory is only valid if you're trying to compute the time required for the ball to return to its initial height of 1.5 m. Unfortunately, you don't know the final height of the ball (when its directly over Julie's head) in this problem.

I would approach this problem in three parts: First, we need to know how long it took the ball to reach the apex of its trajectory at 11 m. For part 2, we need to know the total time it took for the ball to reach Julie. Whatever extra time it took will be time during which the ball is falling, so finally, for part 3, we need to compute how far the ball would fall during that time. That will give you your answer.

Part 1.
There are two ways to calculate how long it takes the ball to reach the apex. We could either use your formula from above knowing the initial and final position of the ball, or we could use the vertical component of the ball's velocity as it leaves Sarah's hand to calculate the amount of time it would take to reach 0 m/s.

First let's use your formula. You know the maximum height the ball achieved (11 m), so you can calculate the time required for the ball to travel from 1.5m to 11m. Incidentally, this will be the same as the time that would be required for the ball to drop from 11m to 1.5m, which is a slightly easier calculation:

$y=y_{initial}+v_{initial} \cdot t + \frac{1}{2} \cdot a \cdot t^2$

$1.5=11+0 \cdot t + \frac{1}{2} \cdot (-9.81) \cdot t^2$

$t = 1.39$s

Now let's use this to calculate the initial vertical velocity of the ball when it left Sarah's hand (this isn't absolutely necessary, but it'll be good to make sure we get the same answer this way as the we do for the second way which we'll get to momentarily).

$v=v_{initial}+a \cdot t$

We know that 1.39s after the ball left Sarah's hand, its vertical velocity was 0.

$0=v_{initial}-9.81 \cdot 1.39$

$v_{initial}=13.65$ m/s

Okay, so now let's try it the second way: As it turns out, the question actually gave you extra information that you also could use to calculate the vertical velocity in a different way. It told you the total velocity of the ball as it left Sarah's hand: 26.75 m/s. If you know that and the horizontal velocity, you can easily compute the vertical velocity because

$v^2=v_h^2+v_v^2$.

So what was the horizontal velocity? Well, remember that the horizontal velocity remains constant throughout the flight of the ball. It's not affected by gravity or vertical velocity. Now recall that the problem tells you that the ball's total velocity was 23 m/s at its maximum height. Since we know the ball's vertical velocity was 0 at the apex of its trajectory, the 23 m/s must have all been in the horizontal direction.

So now we can solve for the vertical velocity:

$26.75^2=23^2+v_v^2$

$v_v=13.66 m/s$

Good! We got the same answer both ways, so clearly we must have done it correctly. And, of course, if we plug this calculation into

$v=v_{initial}+a \cdot t$

we'll get the same answer for t, 1.39 s.

Part 1 complete!

Part 2.
Okay, so now we move on to step 2, where I'll let you do a little more of the work. Your goal here is to figure out how long it takes for the ball to travel from Sarah to Julie (or at least to a point directly above Julie's head).

You know the horizontal velocity of the ball, so all you really need to know is how far apart they're standing (then just use distance=rate x time). In order to calculate the distance between the two girls, you need to go back to the first part of the problem, where Julie threw the ball to Sarah. That part told you the total velocity and the angle of the ball as it left Julie's hand. With that information, it's straightforward to calculate the horizontal and vertical components of the ball's velocity. Once you know the vertical velocity, you can calculate the time required for the ball to travel from Julie to Sarah using your formula. You already know y (1.5 m, the height where Sarah caught the ball), $y_initial$ (also 1.5 m, the height where Julie threw the ball), $v_initial$ ($23 \cdot \sin {33}$), and a. Alternatively, you could calculate the time it took the ball to reach the apex, and then use your trick of doubling it since the trajectory is symmetric in that part of the problem (Sarah caught the ball at the same height as Julie threw it). Either way, once you calculate t, you can use that along with the horizontal component of the velocity to compute the distance between the two girls.

Once you know how far apart Julie and Sarah are standing, you can use the horizontal velocity from the second part of the problem (23 m/s) to calculate how long it will take to travel the full distance between them. That's part 2 of the solution.

Part 3.
Finally, you just need to calculate the final position of the ball after the number of seconds you calculated in part 2. You can either figure out how far the ball dropped during the remaining time after the ball reached the apex (in other words, the total time you computed in part 2 minus the time required to reach the apex that you computed in part 1, i.e. 1.39 s), or you can compute it directly using the initial height and vertical velocity of Sarah's throw and total time of travel that you calculated in part 2. Either way you do it, you'll once again use your formula, and you should get the same answer.

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