Question

if the velocity of projection of projectile is tripled at a given place ,then it's maximum range will be

Answer 1

Recall how to split the initial velocity to its component horizontal and vertical components:

ux=ucosθ;uy=usinθ

The vertical displacement, sy, is taken to be zero to find the time taken for the projectile to hit the ground again.

sy=uyt+12(−g)t2

0=usinθt−12gt2

0=t(usinθ−gt2)

usinθ−gt2=0 or t=0 (rejected)

t=2usinθg

Substituting the time taken for the projectile to hit the ground again into the displacement function in the horizontal direction, sy, we can get the function for the range of projectile.

sx=uxt

sx=ucosθ×2usinθg

sx=u2sin2θg

After all these calculations, we find that the range, sx is proportional to the square of initial velocity, u. (Since both t and θ are constant)

Therefore, doubling the initial velocity will result in the range being 4 times the original.