Question

Ball is thrown from the ground with velocity u at an angle theta with horizontal the horizontal range of the ball on the ground is are

Answer 1

To find horizontal range, only horizontal motion is considered.

Here initial velocity is the horizontal component of u, i.e., u cosθ.

Well, horizontal range is the horizontal displacement of the projectile from the point of projection to the destination point.

We know that displacement = verify velocity × time.

So if the projectile covers the horizontal range in t seconds, is,

R = u cosθ t.

But t, time of projection is,

t = 2 u sinθ / g

This is given by the application of second kinematic equation in vertical motion,

s = u t + (a t²) / 2,

where,

s = 0 (vertical displacement is same for upward and then downward motion),

u = u sinθ (vertical component),

a = - g (acceleration due to gravity is acting vertically downwards).

So,

R = u cosθ · 2 u sinθ / g

R = u² · 2 sinθ cosθ / g

But 2 sinθ cosθ = sin(2θ). So,

R = u² sin(2θ) / g