Question

An object is projected with initial velocity of 100 ms^–1 and angle of 60. Find the vertical velocity when its horizontal displacement is 500 m. (g = 10 ms^–1)

(A) 13.35 ms^–1 

(B) – 13.35 ms^–1 

(C) – 8.65 ms^–1 

(D) 98 ms^–1

Answer 1

Correct option: (B) – 13.35 ms^–1

Explanation:

stone is projected with V₀ = 100 m/s at 60°.

i.e. horizontal velocity = 100 cos 60 = 100 × (1/2) = 50 m/s

vertical velocity = 100 sin 60 = 100 × (√3 / 2) = 50 √3 m/s.

also horizontal displacement = 500 m. horizontal velocity is always constant

t = {(displacement) / (velocity)} = {(500) / 50} = 10 s.

Now when stone will reach top, using V = u + at as V = 0, 0 = V₀¹ + at here

V₀¹ = vertical component of velocity.

i.e. 0 = – 50 √3 + 10t

∴ t = 5 √3 = 8.66 sec

i.e. at 8.66 sec, object is at top. & then in remaining 1.33 sec, it will move to ground.

as V = V₀" + at

V₀" = initial velocity while moving down, V₀" = 0

∴ V = at = – 10 × 1.33 = – 13.3 m/s.