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 =

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

 

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I’m pretty sure it’s i think : /