Question

A projectile is projected with velocity ‘u’ making an angle θ with the horizontal direction. Find: 

1. Time of flight

2. Horizontal range

Answer 1

1. Let H be the maximum height reached by the projectile in time t₁

For vertical motion,

The initial velocity = usin θ

The final velocity = 0

Acceleration = - g

using, v² = u² + 2as

0 = u² sin² θ - 2gH

2gH =u²sin²θ

u² sin² θ

H = \(\frac {u^2 sin^2\theta}{2g}\)

2. Let t, be the time taken by the projectile to reach the maximum height H.

For vertical motion,

initial velocity =usinθ

Final velocity at the maximum height = 0

Acceleration a = - g

Using the equation v = u + at₁

0 = usin θ - gt₁

gt₁ = usin θ

t₁ = \(\frac {usin \theta}{g}\)

Let t₂ be the time of descent.

But t₁ = t₂

i.e. time of ascent = time of descent

Time of flight T = t₁ + t₂ = 2t₁

_{∴ T = \(\frac{ sin \theta}{g}\)}.