Question

A body of mass m, projected at an angle of θ from the ground with an initial velocity of v, acceleration due to gravity is g, what is the maximum horizontal range covered?
a) R = v2 (sin 2θ)/g
b) R = v2 (sin θ)/2g
c) R = v2 (sin 2θ)/2g
d) R = v2 (sin θ)/g​

Answer 1

Range of Projectile

• Let's assume that a body of mass m is projected at an angle of θ from the ground with an initial velocity of v.

We know that time period of projectile is:

$T = \dfrac{2v\sin( \theta) }{g}$

Now, range (R) will be :

$R = u_{x} \times T$

$\implies R = v \cos( \theta) \times \dfrac{2v \sin( \theta) }{g}$

$\implies R = \dfrac{2 {v}^{2} \sin( \theta) \cos( \theta) }{g}$

$\implies R = \dfrac{ {v}^{2} \{2 \sin( \theta) \cos( \theta) \} }{g}$

$\boxed{ \implies R = \dfrac{ {v}^{2}\sin(2 \theta) }{g} }$

OPTION a) IS CORRECT ✔️

Hope It Helps.