H/w help thx a bunch

Question

If an arrow is shot straight up from the ground level on the moon with an initial velocity of 100 feet per second, its height will be given by: h(t)= -2.65t^2 +100t feet after t seconds

Find a function for the velocity of the arrow at time t

v

What is the maximum height of the arrow?? [Hint: It happens when the velocity is zero.]

Maximum Height 

What is the velocity when it hits the ground?

 __________________feet per second

Answer 1

Given: h(t)= -2.65t² +100t

(1) Find a function for the velocity of the arrow at time t.

v(t) is the same as the derivative h(t).

v(t) = h'(t) = -5.3t + 100

(2) What is the maximum height of the arrow?

It means when v(t) = 0

0 = -5.3t + 100

5.3t = 100

t = 1000/53

plugin the value of t in h(t)

h(1000/53)= -2.65(1000/53)² +100(1000/53)

=943.40 feet

(3) What is the velocity when it hits the ground?

it mean h(t) = 0

0 = -2.65t² +100t

0 = t(-2.65t + 100)

t = 0, -2.65t + 100 = 0

Eliminate t=0

-2.65t = -100

t = 100/2.65 =2000/53

plugin t in v(t)

v(2000/53) = -5.3(2000/53) + 100

v(2000/53) = -100 ft./ sec.