Question

I am throwing a ball on the groundfrom the 12th floor of abuilding. The height, in meters, of the of the ball is above the ground t seconds after being thrown is modeled by
h(t)=−2t^2+4t+30


(a) When will the ball reach its maximum height?

(b) What is the height of my building?

Answer 1

You should proof-read your post before submitting it, as it stands it is
well garbled.

First of all, what planet is this building on, since h(t) = -2t^2 + 4t + 30
does not fit metres and seconds.

The 4t + 30 would tell me you are throwing this upwards from a height of
30 metres with an intial velocity of 4 m/s

so the height of the building would be 30 m
h '(t) = -4t + 4
at a max height, this is zero, so
-4t+4=0
t = 1

h(1) = -2(1^2) + 4(1) + 30 = 32 metres

the max height is 30 m after 1 second