Question

for a given angle of projection if the time of flight of projectile is doubled then the horizontal range will increase to

Answer 1

ind a relation between time of flight T and maximum height H and then it is easy.

H=u2sin2θ2g … … (1)

T=2usinθ2g

T2=4u2sin2θg2 … … (2)

Divide equation (1) by equation (2) to get

H=T2g8

This makes H∝T2

there fore it will be 4 times

Answer 2

We know the time of flight t = 2Vyo/g where Vyo is the vertical component of the launch velocity. We know the range R = Vxo*t = Vxo*2Vyo/g = Vo*cosθ*2*Vo*sinθ/g = 2Vo²/g * cosθ*sinθ = 2Vo²/g *(1/2)*sin2θ = Vo²/g *sin2θ 
If the launch angle is constant but the range is doubled then Vo² must be doubled or Vo' = Vo*√2 so the launch velocity is √2 the original launch velocity which means the time of flight is √2 the original flight time

Therefore the answer is 4 times

Hope it helps you
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