Question

A ball is thrown into the air with an upward velocity of 32 ft/s. Its height h in feet after t seconds is given by the function h = –16t2 + 32t + 6. a. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. b. What is the ball’s maximum height?

Answer 1

@Study2Learn

Answer 2

I got 2 seconds, 6 feet, is that correct?

Answer 3

You may be right though, I don't want to get you wrong.

Answer 4

I believe it would be 1 second, 22 ft though.

Answer 5

@mhchen pls check this whenever you're online

Answer 6

complete the square:

\(h(t) = –16t^2 + 32t + 6 \qquad \equiv \color{red} {22 - 16 (t-1)^2
} \)

How big can that bit in red get?! And when does it get there?

Answer 7

@TheSmartOne

Answer 8

@TheSmartOne are you still there?

Answer 9

You just need to find the maximum.

t = -b/2a
That will give you the maximum x-coordinate which is the the time at which the ball is at maximum height

Plugging in that t-value, and solving for h will give you the maximum height

Answer 10

I did that before, and I got 2 seconds, 6 ft. is that correct?

Answer 11

No

The equation is in the form of
ax^2 + bx + c

so do -b/2a
You take the coefficients