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 = v^2 (sin 2θ)/g

(b) R = v^2 (sin θ)/2g

(c) R = v^2 (sin 2θ)/2g

(d) R = v^2 (sin θ)/g

I have been asked this question in a job interview.

My question is based upon Simple Projectile Motion in chapter Motion in a Plane of Physics – Class 11

Answer 1

Correct choice is (a) R = v^2 (sin 2θ)/g

Explanation: The calculation for range uses the horizontal component of the initial velocity and time. Range = distance x Time taken for the motion. The time can be found by utilising the fact that the vertical velocity becomes zero at the maximum height (i.e. at half of the time period) and using the first equation of motion for this. Hence, on calculating, the range will be obtained as R = v^2 (sin 2θ)/g.