Question

. A ball is thrown upward. Its height (h, in feet) is

given by the function h = 16t^2+ 64t + 34, where t is
the length of time (in seconds) that the ball has been
in the air. What is the maximum height that the ball
reaches?
A. 3 ft B. 51 ft C. 63 ft D. 67

Answer 1

64 feet

Explanation:

You can solve this by completing the square and turning the equation into that of a parabola, where the uppermost vertex is the highest point in the ball's flight path.

`h=-16(t^2-4t+4)+98`

`h=-16(t-2)^2+98`

Since the first term there is negative, the largest possible answer that you can get is if that term is 0. The only way to make it 0 is for t to be 2, so you know that that is when the highest point is. If the first term is 0, all that is left is 98, which is the highest height. However, since you initially start at 98, you only end up going 64 feet up (I guess C is the closest to that?). Hope this helps!

Answer 2

Well none of the answers are correct as maximum height is found by doing -b/2a which gives the x coordinate of the vertex also known as the heighest or lowest point. But when plugged in it gives the lowest point assuming you made the mistake of making the 16 positive when its supposed to be negative you would plug it in and get 98 which isn't an answer choice assuming 34 is the ground when you subtract that you get 64 whch still isn't an answer choice check your numbers in the equation