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I'm confused about the following problem.

A ball is shot from the ground into the air. At a height of $9.1\text{ m}$, its velocity is $v = (7.6\hat{\imath}+ 6.1\hat{\jmath})\text{m/s}$, with $\hat{\imath}$ horizontal and $\hat{\jmath}$ upward.

  1. To what maximum height does the ball rise?
  2. What total horizontal distance does the ball travel?
  3. What is the magnitude and angle (below the horizontal) of the ball’s velocity just before it hits the ground?

The given answers are: (1) 11m, (2) 22.76m, (3) 16.57m/s, 62.69

We know $S=ut + 1/2 at^2$. So,

$$Y = Y_0 + V_{0y}t - 1/2gt^2$$

and,

$$X=X_0 + V_{0x}t$$

and $v=u+at$, so

$$V_x = V_{0x} - gt$$

and

$$V_y = V_{0y} - gt$$

and we're given that $V_x=7.6$ and $V_y=6.1$

I can realize that here $Y = 9.1\text{ m}$ but I cant find the $V_0$ and also the initial angle. How can I do that?

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Welcome to Physics! Please see our homework policy. We expect homework problems to have some effort put into them, and deal with conceptual issues. If you edit your question to explain (1) What you have tried, (2) the concept you have trouble with, and (3) your level of understanding, I'll be happy to reopen this. (Flag this message for ♦ attention with a custom message, or reply to me in the comments with @Manishearth to notify me) –  Manishearth Feb 1 '13 at 17:13
    
@Manishearth: Ive tried to explain where im facing problem. Would you pls consider it opening again? –  Samiron Feb 1 '13 at 18:01
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Hi Samiron! I've edited your question, and in particular the title, to focus it on what it seems like you're trying to ask. Please check if this is accurate. –  David Z Feb 1 '13 at 18:51

1 Answer 1

up vote 1 down vote accepted

It may help you to think of the ball being thrown at the given velocity from the top of a cliff 9.1 m high. Also, recall that since i and j are orthogonal, they are independent, so you can split the problem into x and y sub-problems.

This solution is a bit naive, but it is intentionally so.

STEp 1:

How long does it take for the object to slow down to Vy=0?

0 = 6.1 - 9.8 t -- Solve for t.

Now that you have t from the top of the cliff to the top of the arc, you can calculate the height. How far will a ball with initial velocity 6.1 m/s rise in t seconds?

You can add that height to 9.1 to get Ymax in terms of the question axis. -- At this point we can dispense with the cliff idea. It's served it's purpose.

STEP 2: Now that we know the ground -- Maximum distance, we can figure out how long the ball is in the air. Simply figure out how long it would take it to fall from that height.

We know tup = tdown, so total flying time is twice as long as it would take to fall. Lets call it ttotal.

The horizontal distance is simply 7.6 m/s * ttot.

Step 3: We know the i component, all we need is the j component for Vfinal. Simple drop the ball from Ymax and figure out how fast it's going. Then apply some basic trig to figure |V| and the angle.

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Thanks a lot.. i will try this way and get back with a reply :) –  Samiron Feb 2 '13 at 3:20

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