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Question

Physics Question on projectile motion

What is the relationship between time of flight TT and horizontal range RR ? (where θ\theta is angle of projection with the horizontal)

A

R=gTtan θR=\frac{gT}{tan\, \theta}

B

R=gT22 tan θR=\frac{gT^{2}}{2\,tan\, \theta}

C

R=gT2tan θR=\frac{gT^{2}}{tan\, \theta}

D

R=gT2 tan θR=\frac{gT}{2\,tan\, \theta}

Answer

R=gT22 tan θR=\frac{gT^{2}}{2\,tan\, \theta}

Explanation

Solution

Horizontal range R=u2 sin 2θgR=\frac{u^{2}\,sin\,2\theta}{g} Time of flight T=2u sin θgT=\frac{2u\, sin\,\theta}{g} where u is the velocity of the projection and θ\theta is the angle of projection with the horizontal R=u2 sin 2θgR=\frac{u^{2}\,sin\,2\theta}{g} or R=2u2 sin θ cos θgR=\frac{2u^{2}\,sin\,\theta\,cos\,\theta}{g} or R2 cos θ=u2 sin θg…(i) \frac{R}{2\,cos\,\theta} = \frac{u^{2}\,sin\,\theta}{g} \ldots\left(i\right) T=2u sin θgT=\frac{2u\,sin\,\theta}{g} or T2=4u2 sin2 θg2T_{2}=\frac{4u^{2}\,sin^{2}\,\theta}{g^{2}} T2=4g(u2 sin θg)sin θT^{2}=\frac{4}{g}\left(\frac{u^{2}\,sin\,\theta}{g}\right)sin\,\theta =4g(R2 cos θ)sin θ=\frac{4}{g}\left(\frac{R}{2\,cos\,\theta}\right)sin\,\theta (Using (i)) or T2=2Rgtan θT_{2} =\frac{2R}{g} tan\,\theta or R=gT22 tan θR=\frac{gT^{2}}{2\,tan\,\theta}