Question

Derive an expression for the maximum height reached, time of flight and range of a projectile.

Answer 1

(i) Let H be the maximum height reached by the projectile in time t₁ For vertical motion,

The initial velocity = u sinθ

The final velocity = 0

Acceleration = - g

∴ using, v² = u² + 2as

0 = u²sin² θ - 2gH

2gH = u²sin² θ

H = \(\frac {u^2 sin^2 \theta}{2g}\)

(ii) Let t, be the time taken by the projectile to reach the maximum height H.

For vertical motion,

initial velocity = usin θ

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 = \(\frac {2usin \theta}{g}\)

(iii) Let R be the range of the projectile in a time T. This is covered by the projectile with a constant velocity ucos θ.

Range = horizontal component of

velocity × Time of flight

i.e, R = ucos θ. T

R = ucos θ. \(\frac{2usin\theta}{g}\)

R = \(\frac{u^2sin2\theta}{g}\)

∵ 2 sin θ.cos θ = sin2 θ.