Question

Calculus Question, homies <3
If a ball is thrown vertically upward from the roof of a 32 foot tall building with a velocity of 16 ft/sec, its height after t seconds is s(t) = 32 + 16 t - 16 t^2.
What is the maximum height the ball reaches?
What is the velocity of the ball when it hits the ground (height 0)?

Answer 1

1) You may want to take derivative then set it equal to 0 then solve for t.



We can call this t "\(t_{top}\)"

2) You would need to find \(\left.\dfrac{ds(t)}{dt}\right|_{~t=2t_{top}}=v(2t_{top})\)

Since it take \(t_{top}\) for ball to reaches maximum height and another \(t_{top}\) for ball to hit ground, hence \(2t_{top}\)

Does that make sense?

Answer 2

Sorry, I was in between classes.

So the derivative of what you were saying is v(2ttop)? Where does the 32 feet come in to play?

Answer 3

If I take the derivative of that equation, it would be -32t+16?

Answer 4

I was able to figure this one out.
First, I got the derivative of s(t)=16t-32t^2+32. so s'(t)=16-32t. I solved for t and ended up getting .5. I put .5 in to the equation 16t-32t^2+32 and ended up getting 36 for the answer

For the second answer, I set 16t-32t^2+32 equal to 0 and solved. I ended up getting -1 and 2. I plug 2 into the derivative and got -48!