Question

If time taken by the projectile to reach Q is $T$, then PQ $=$

• A. $Tvsin\theta$
• B. $Tvcos\theta$
• C. $Tvsec\theta$
• D. $Tvtan\theta$

Answer 1

Answer: (D)

$Tvtan\theta$

Range: $R=\frac{v^2\sin 2(90-\theta )}{g}=\frac{v^2\sin 2\theta }{g}$

$PQ=\frac{R}{\cos \theta }$

$T=\frac{2v\cos \theta }{g}$
Combining $PQ=Tv\tan \theta$