Question

A projectile is sitting on top of a cliff that is 250m above ground. The projectile is

launched with a horizontal speed of 25m/s

(a) How far from the base of the cliff did the projectile “land?”

(b) What is the vertical speed of the projectile when it lands [just before it hits
the ground]?

(c) How long did it take the projectile to reach the ground?

Answer 1

a. 178.5 m

b. 69.97 m/s

c. 7.14 s

Further explanation

Given

height = 250 m

vox = 25 m/s

Required

distance from base

vertical speed(vy)

time to reach ground

Solution

voy = 0

h=voy.t + 1/2 gt²

h = 1/2gt²

250 = 1/2.9.8 t²

t=7.14 s

x = vox.t

x = 25 x 7.14

x = 178.5 m

vy = voy + gt(voy=0)

vy = 9.8 x 7.14

vy = 69.97 m/s