Answer:
Range will become 4 times of initial range
Explanation:
Let the velocity of projection is u
And angle at which projectile is projected is [tex]\Theta[/tex]
And acceleration due to gravity is [tex]g\ m/sec^2[/tex]
So range of projectile is equal to [tex]R=\frac{u^2sin2\Theta }{g}[/tex]........eqn 1
Now in second case it is given that velocity of launching is doubled
So new velocity [tex]u_{new}=2u[/tex]
So new range will be equal to [tex]R_{new}=\frac{(2u)^2sin2\Theta }{g}=\frac{4u^2sin2\Theta }{g}[/tex] .....eqn 2
Now dividing eqn 2 by eqn 1
[tex]\frac{R_{new}}{R}=\frac{4u^2sin2\Theta }{g}\times \frac{g}{u^2sin2\Theta }[/tex]
[tex]R_{new}=4R[/tex]
So if we double the initial launch speed then range will become 4 times
