Question

A missile is launched from the ground making 45 degree with the horizontal to hit a target at a horizontal distance of 300 km if it is required to hit a target at a horizontal distance of 675 km launched at same angle with horizontal find the percentage change in its velocity of projection.

Answer 1

The range is
R=v^2/g sin⁡2θ
sin⁡2θ=sin⁡2(45)=1
R^'/R=(v^'/v)^2
(v^'/v)=√(R^'/R)
(v^'/v)-1=√(R^'/R)-1=√(675/300)-1=0.5.
Therefore, the percentage change in its velocity of projection is 50%.