## anonymous 3 years ago a batting machine launches a ball straight up in the air at a velocity of 100 ft/s. the machine is 65 feet tall what is the maximum height the ball can attain how fast will it be moving at this point?

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1. anonymous

Well, let's start from the top. It's important to be able to derive the equations you need for physics without relying on memorization. Do you know calculus?

2. anonymous

Yes

3. anonymous

Good. Assume that the ball, at any given moment, has constant acceleration $$a$$. First, find a formula for $$v(t)$$, the vertical velocity of the ball at any given time.

4. anonymous

I know that it needs to involve a (which would be negative due to gravity) and x over a certain period of time t. So $v(t)=a(\int\limits_{0}^{t}xdt)$?

5. anonymous

Well don't worry about what the value of $$a$$ yet (although you are right in that it is negative and the gravitational acceleration constant, which is $$-g = -9.8 \text{m}/\text{s}^2$$. Solving for $$v(t)$$ should be easy. If the derivative (acceleration) is constant, then $$v(t) = a t + C$$ right?

6. anonymous

Ugh! Yes. It has been so long since I have done any word problems like this and it frustrates me that I can't remember how to start them off. I am really trying to help my cousin with her homework :/

7. anonymous

Ok. well, we still have on more time-integral left. since this isn't for you, I'll just give you the solution: $$y = y_0 + v_0 t + a t^2/2$$ I trust you can find approp. values for the constants, right?

8. anonymous

$y_{0}=65 ft v _{0}=100ft/s$

9. anonymous

Thank you so much for the help. I am clearly too tired after my own exams to be helping anyone else.

10. anonymous

Yes @ your values. Max height can be found using single-derivative optimization (find where $$v(t) = 0$$).