Question

Show and explain work for the following problem:

10. From 4 feet above a swimming pool, Deborah throws a ball upward with a velocity of 32 feet per second. The height, h(t), of the ball t seconds after Deborah throws it and is given by the equation: h(t) = -16t^2 + 32t + 4. Find the maximum height reached by the ball and the time this height is reached.

Answer 1

maximize h(t). find 't' for which the derivative of the given function is zero. here you get t = 1. giving maximum height reached by the ball as 36 ft. Alternatively, h(t) is a quadratic expression whose max is given by f(-b/2a) when f(x) = \[ax ^{2}+bx+c\]

Answer 2

is that the answer to the question?

Answer 3

No, we don't give out answers here, just show you how to work a problem.

Answer 4

yes I am well aware of that

Answer 5

Well, this is a quadratic, so you can use \[-\frac{ b }{ 2a }\] to get time at max height, then plug into original equation to get actual max height.

Answer 6

@Aani Is that the correct answer to the problem? Cause I want to make sure I have the right answer and I understand what to do

Answer 7

@Australopithecus Who's more right???

Answer 8

@Firejay5 yes the answer is 32.

Answer 9

Thank you so much! :D

Answer 10

I don't think it's 32, is it?

Answer 11

*36 ft

Answer 12

I plugged 1 back into the original equation and got 20

Answer 13

Sorry, err , it's 20 ft.