Question

Maximum range of projectile is 10m what will be its maximum range if its velocity of projection is doubled

Answer 1

Maximum range of a projectile is 10m

i.e R= v^2 sin2Ф/g= 10m

v is the velocity of projection and Ф the angle of projection

if velocity of projection is doubled then range becomes four times according to the equation

R2 = (2v)^2 sin2Ф/g= 4R

 maximum range = 4*10=40m

Answer 2

Given :

Maximum range of projectile = 10 m

To Find :

Its maximum range when its velocity of projection is doubled =  ?

Solution :

∴we know that formula for Range of a projectile with velocity of projection u and angle of projection $\theta$ is given as :

$R = \frac{u^{2}sin2\theta }{g}$

So from the above formula we can see that the Range is directly proportional to the square of the velocity of projection , i.e. :

R   $\alpha$    $u^{2}$

So , according to this if the velocity of projection is doubled so the range will be 4 times the initial range , i.e.

R'  = $4\times R$

   = 4 $\times$ 10 m

   = 40 m

Therefore after doubling the initial velocity the new maximum range is 40 m .