If an arrow is shot upward from the surface of the moon with a velocity of 56 meters per second, its height (in meters) above the moon t seconds later is given by H(t) = 56t – 0.83t2 .

Question

a) Find the velocity of the arrow after 1 second.

b) Find the time when the arrow lands back on the moon.

c) Find the velocity of the arrow when it hits the moon.

Answer 1

velocity = the derivative of H(t) = H'(t) = 56-1.66t

H'(1) = 56-1.66 = 54.34 meters per second

0=H(t) = 56t-0.83t^2= t(56-0.83t)

t= 0, 56/.83

v(56/.83) = H'(56/.83)= 56-1.66(56/.83)