Kinematics

JEE Main level, single correct, medium

248 of 308

Question 248 of 308

A projectile is projected with velocity of 25ms−125m{s}^{-1} at an angle θ\theta with the horizontal. After tt seconds its inclination with horizontal becomes zero. If RR represents horizontal range of the projectile, the value of θ\theta will be : [use useg=10ms−2useg=10m{s}^{-2}]

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Answer: (D) cot⁡−1(R20t2){\cot }^{-1}\left(\dfrac{R}{20{t}^{2}}\right)

From the information given in the question, after time tt the inclination of the projectile becomes zero i.e. the projectile is at the highest point. Therefore,
t=T2=usin⁡θgt=\dfrac{T}{2}=\dfrac{u\sin \theta }{g}.
The rang of the projectile R=2u2sin⁡θcos⁡θgR=\dfrac{2{u}^{2}\sin \theta \cos \theta }{g}.
Now,
Rt2=2u2sin⁡θcos⁡θg×g2u2sin⁡2θ=2gcot⁡θ⇒θ=cot⁡−1(R20t2)\dfrac{R}{{t}^{2}}=\dfrac{2{u}^{2}\sin \theta \cos \theta }{g}\times \dfrac{{g}^{2}}{{u}^{2}{\sin }^{2}\theta }=2g\cot \theta \Rightarrow \theta ={\cot }^{-1}\left(\dfrac{R}{20{t}^{2}}\right)

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