Question

The range of projectile remains the same for the angle of projections 30° and 60°

Answer 1

Let u be the initial velocity of projectile and θ be the angle of projection.

To find horizontal range of a projectile we need to consider its horizontal motion.

• Initial horizontal velocity, $\sf{u_x=u\cos\theta}$

Since no horizontal external forces are acting on the projectile,

• Horizontal acceleration, $\sf{a_x=0\ m\,s^{-2}}$

Horizontal range of the projectile is nothing but the horizontal displacement of the projectile.

Since displacement is the product of velocity and time, horizontal range,

$\longrightarrow\sf{R=u_x\cdot t}$

$\longrightarrow\sf{R=u\cos\theta\cdot t}$

Here t is the time of flight of the projectile and is given by,

$\longrightarrow\sf{t=\dfrac{2u\sin\theta}{g}}$

which can be derived by considering vertical motion of the projectile.

Hence,

$\longrightarrow\sf{R=u\cos\theta\cdot\dfrac{2u\sin\theta}{g}}$

$\longrightarrow\sf{R=\dfrac{u^2\cdot2\sin\theta\cos\theta}{g}}$

We know that,

• $\sf{2\sin\theta\cos\theta=\sin(2\theta)}$

Hence,

$\longrightarrow\sf{R=\dfrac{u^2\sin(2\theta)}{g}}$

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If $\sf{\theta=30^o,}$

$\longrightarrow\sf{R=\dfrac{u^2\sin(2\times30^o)}{g}}$

$\longrightarrow\sf{R=\dfrac{u^2\sin60^o}{g}}$

$\longrightarrow\sf{R=\dfrac{u^2\sqrt3}{2g}}$

If $\sf{\theta=60^o,}$

$\longrightarrow\sf{R=\dfrac{u^2\sin(2\times60^o)}{g}}$

$\longrightarrow\sf{R=\dfrac{u^2\sin120^o}{g}}$

$\longrightarrow\sf{R=\dfrac{u^2\sqrt3}{2g}}$

Hence the range of projectile remains same for the angles of projections 30° and 60° and for same initial speed.