Question

If an arrow is shot upward on the moon with velocity of 47 m/s, its height (in meters) after t seconds is given by

h(t)=47t_0.83t2.


(a) Find the velocity of the arrow after 8 seconds.
Velocity =
(b) Find the velocity of the arrow after a seconds.

Velocity =
(c) When will the arrow hit the moon?

Time =
(d) With what velocity will the arrow hit the moon?

Velocity =

Answer 1

h(t) = 47t - 0.83t^2

Vo = 47m/s

V (t) = 47 - 1.66 t

a) t = 8 s; V(8) = 47 - 1.66(8) = 33.72 m/s

b) t = a; V(a) = 47 -1.66(a)

c) Hit the moon t =?

h(t) = 0 => 47t - 0.83 t^2 = 0

Factor: t (47 - 0.83t) = 0

47 - 0.83t = 0 => 0.83t = 47

t = 47 / 0.83 = 56.63 s

d) Velocity when hit the moon

The velocity of the arrow when it hits the moon is the same initial speed with opposite sign.

V = - 47 m/s