Question

Obtain the equation for projectile and show that the range is maximum at 45°

Answer 1

Here,

There will be two types of Motion will occur in Projectile.

Motion along horizontal direction.

The velocity of the body in horizontal direction is constant, so the acceleration ${a}_{x}$ in horizontal direction is zero.

x = ${x}_{0}+{u}_{x}t+\dfrac{1}{2}{a}_{x}{t}^{2}$

But, ${x}_{0}=0$, ux= uCosØ ${a}_{x}=0$

x = uCosØ t

t = x/uCosØ ------(1)

Again,

Motion along vertical direction,

y = uSinØ t + 1/2(-g)t²

y = uSinØ t - 1/2 gt² -----(2)

Putting the value of 't' in eq(2)... we get,

y = uSinØ (x/uCosØ) - 1/2g[x/uCosØ]²

y = x tanØ - gx²/2u² Cos²Ø

For maximum Range.

Given :-

Ø = 45°

As we know that,

R = u²Sin2Ø/g

R = 2u² Sin45° × Cos45°/g

R = 2u² (1/√2)(1/√2)/g

R = 2u²/2g

R = u²/g

Hence, Proved that Range will be maximum when Ø = 45°.

For same value of initial velocity, horizontal range of body is same for complementary angles.

For Instance :-

${R}_{30^{\circ}} = {R}_{60^{\circ}}$

or,

${R}_{20^{\circ}} = {R}_{70^{\circ}}$