Kinematics

JEE Main level, single correct, medium

262 of 308

Question 262 of 308

The trajectory of a projectile near the surface of the earth is given as y=2x−9x2.y=2x-9{x}^{2}. If it were launched at an angle θ0{\theta }_{0} with speed v0{v}_{0} then (g=10ms−2):\left(g=10{ms}^{-2}\right):

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Answer: (A) θ0=cos⁡−1⁡(15){\theta }_{0}={\cos }^{-1}⁡\left(\dfrac{1}{\sqrt{5}}\right) and v0=53ms−1{v}_{0}=\dfrac{5}{3}m{s}^{-1}

y=2x−9x2y=2x-9{x}^{2}
Comparing above equation with general equation of trajectory of projectile y=xtan⁡⁡θ(1−xR)y=x\tan ⁡\theta \left(1-\dfrac{x}{R}\right)
tanθ=2tan\theta =2
⇒sinθ=25\Rightarrow sin\theta =\dfrac{2}{\sqrt{5}} or cosθ=15cos\theta =\dfrac{1}{\sqrt{5}}
θ=sin−1(25)\theta ={sin}^{-1}\left(\dfrac{2}{\sqrt{5}}\right) or θ=cos−1(15)\theta ={cos}^{-1}\left(\dfrac{1}{\sqrt{5}}\right)
And R = 2/9
R=v02sin⁡⁡2θg=29R=\dfrac{{{v}_{0}}^{2}\sin ⁡2\theta }{g}=\dfrac{2}{9}
After substituting the value for g and sin 2θ. We get v0=35{v}_{0}=\dfrac{3}{5} m/s

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