Question

A projectile is launched at an initial angle ?₀ above the horizontal from the edge of the top of a cliff
that is 15 m high. It hits the ground below, at the level of the base of the cliff, 230 m from the cliff,
after a flight time of 4.2 s.
a) What was the initial horizontal component of the projectile’s velocity?
b) What was the initial vertical component of the projectile’s velocity?
c) What were the initial angle ?₀ and initial speed ?₀?
d) What was the projectile's speed just before it struck the ground?

Answer 1

The vertical velocity would v_osinθ  and the horizontal velocity would be v_ocosθ

A)

Initial horizontal component of velocity

Total horizontal distance travelled

R = 230 m

Time taken to travel this distance

t = 4.2 s

R/t = v_ocosθ

V_ocosθ = 230/4.2

V_ocosθ = 54.8 m/s

B)
Initial vertical component of velocity

Total vertical distance travelled

H = ut - ½gt²

15 = (v_osinθ x 4.2) - 0.5 x 9.8 x 4.2

V_osinθ =35.58/4.2

v_osinθ = 8.47 m/s

C)

Initial angle

V_osinθ/V_ocosθ = tanθ

Tanθ = 35.58/230

θ = 8.79^o

Initial speed

V_o² = (v_osinθ)² + (V_ocosθ)²

V_o² = (35.58/4.2)² + (230/4.2)²

V_o = 55.4 m/s

D)

V = u - gt

V = 8.47 - 9.8 x 4.2

V = -32.7 m/s ( The minus sign just indicates that the final velocity is downwards)