Question

4. From the top of a building 45 m above the ground an arrow thrown horizontally with a speed of 250 m/s. a) How long until the arrow reaches the ground? b) At what distance from the building the arrow strikes the ground? c) What is the velocity of the arrow as it strikes the ground?

Answer 1

H = 1/2 g t^2 where t is the time to fall 45 m

t = (2 * H / g)^1/2 = (90 / 9.80)^1/2 = 3.03 sec

b) Sx = Vx * t = 250 m/s * 3.03 s = 758 m

c) V = (Vx^2 + Vy^2)^1/2

Vy = 9.80 m/s^2 * 3.03 /s = 27.7 m/s

V = (250^2 + 27.7^2)^1/2 = 252 m/s

tan θ = 27.7 / 250 = .110

θ = 6.3 deg below the horizontal