Solveeit Logo

Question

Physics Question on Motion in a plane

If the range and time period of projectile are related to R=5T2,R=5{{T}^{2}}, then the angle of projection will be

A

π/2\pi /2

B

π/3\pi /3

C

π/4\pi /4

D

π/6\pi /6

Answer

π/4\pi /4

Explanation

Solution

T=2usinθgT=\frac{2u\sin \theta }{g} R=u2sin2θgR=\frac{{{u}^{2}}\sin 2\theta }{g} According to question, R=5T2R=5{{T}^{2}} u2sin2θg=5(2usinθg)2\frac{{{u}^{2}}\sin 2\theta }{g}=5{{\left( \frac{2u\sin \theta }{g} \right)}^{2}} 2u2sinθ.cosθg=20u2sin2θg\frac{2{{u}^{2}}\sin \theta .\cos \theta }{g}=\frac{20{{u}^{2}}{{\sin }^{2}}\theta }{g} cosθ=10sinθg2\cos \theta =\frac{10\sin \theta }{{{g}^{2}}} tanθ=1\tan \theta =1 or θ=π4\theta =\frac{\pi }{4}