A bit more physics help

Question

A crossbow arrow is accidentally shot straight up into the air from the ground with a velocity of 52 m/s.

a) How long is the arrow in the air?

b) What will be the maximum height of the arrow?

c) How fast will the arrow be traveling 7 seconds after it has been shot out of the crossbow? Will it be going up or down? Support your answer.

Answer 1

As we know:

T_upward = T_downward = V₀/g → T_total = 2*T_upward = 2(V₀/g) → T_total = 2(52/9.8) = 10.61 sec

H_max = V₀²/(2*g) → H_max = (52)²/(2*9.8) = 137.96 m

V = -gt + V₀ → V = -(9.8)(7.0) + 52 = -16.6 m/s (arrow reached its max height and came back toward the earth)

(going down)