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$1.$ A ball is hit with an initial horizontal velocity of 33 ft/sec and initial vertical velocity of 51 ft/ sec. How long does it take for the ball to go from its max height to the ground?

My attempt: I assume I only need to worry about vertical velocity to solve this and gravity is -32 ft/sec. Thus I determined:

$\int -32 dt = v_y = -32t + c$

Given the initial vertical velocity of 33, then $v_y = -32t + 33$. Then to get vertical distance we integrate again which gives us $-16t^2 + 33t = d_y$ (y distance). Plug $0$ in for $d_y$ and you get $t=1.44$. I'm not sure if this is the correct time though. Is this just the total time the ball is in flight?

$2.$ A ball takes 3 seconds to travel from its max height to the ground. The initial horizontal velocity of the ball is 27 m/s.

a. What is the angle at which the ball takes off?

b. What is the initial vertical velocity of the ball?

If I could find the initial vertical velocity, I could do $arctan(v_{yo}/v_{xo})$, but not sure how to find it. Thanks for any help :)

Drake
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3 Answers3

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For $1$:

yes, that's the total time it's in flight. (Also, you should have a definite integral:

$$v_f - v_i = \int_{t_i}^{t_f} -32~dt = -32t \implies v_f = -32t + v_i = -32t + 33$$

For $2$:

You're right about finding the initial vertical velocity first. You know that $-32 t + v_{y,i} = 0$ when $t = 3$ (because at its apex, there is no vertical motion), so can you find $v_{y,i}$?

I suppose you could also model it as a ball fired from a certain (unknown) height, and than after $3$ seconds, you declare that to be the ground. Then find the angle it's moving at, but that's not as natural a solution. (Especially because either way, you compute $v_{y,i}$ first.)

Henry Swanson
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  • For 2: shouldn't it be $-32t + v_{y,i}=0$ since we are dealing with velocity equation? – Drake Dec 10 '13 at 01:21
  • Whoops, I thought you had to find the height as well. Yeah, you can just solve $-32 t + v_{y,i} = 0$ and $t = 3$ to find the initial velocity. – Henry Swanson Dec 10 '13 at 01:38
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I don't see your $1.44$. The travel time is $\frac{33}{16}$. From this we subtract the time take to reach maximum height, which can be computed. However, symmetry tells us this is half the time, so our required time is $\frac{33}{32}$.

For the second problem, think of the calculation in the first problem. By analyzing what went on, you can find what the initial speed should be to give a $6$ second total travel time. From this initial speed and the given horizontal component of it, you can find the angle. The cosine will be involved.

André Nicolas
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For the first part, you're correct that it's the total time in flight. The time you're looking for (from the maximum height to the ground) is half that. (And the units are seconds, of course.)

The correct answer is

$$v = v_0 - at \to t = \frac{v - v_0}{a} = \frac{51-0}{32} \approx 1.59 s.$$

Note: $v_{final} = -v_{initial}$.

For the second part, the horizontal velocity doesn't change. The vertical velocity when it hits the ground will be $v = v_0 - at = 0 - 32 \centerdot 3 = 96$ ft/sec down. The initial vertical velocity is then $96$ ft/sec up.

You have the correct expression to calculate the initial angle.

(Note: I solved part B first!)

John
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