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?

Question

anonymous · Answer

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?

anonymous · Answer

Yes

anonymous · Answer

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.

anonymous · Answer

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 $$v(t)=a(\int\limits_{0}^{t}xdt)$$?

anonymous · Answer

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?

anonymous · Answer

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 :/

anonymous · Answer

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?

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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