Verified Solution • 2025–26 Session moderate

QJEEM19Ph7a7b: JEE Main 2019 Physics - Class 11 JEEM19Ph7a7b Solution: The trajectory of a projectile near the surface of the earth is …

The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x2 . If it were launched at an angle θ\theta 0 with speed v0 then (g = 10 ms–2) :

Multiple Choice Options

  • A θ0=cos⁡−1(15){\theta _0} = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right) and v0=53{v_0} = {5 \over 3} ms-1
  • B θ0=cos⁡−1(25){\theta _0} = {\cos ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right) and v0=35{v_0} = {3 \over 5} ms-1
  • C θ0=sin⁡−1(25){\theta _0} = {\sin ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right) and v0=35{v_0} = {3 \over 5} ms-1
  • D θ0=sin⁡−1(15){\theta _0} = {\sin ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right) and v0=53{v_0} = {5 \over 3} ms-1

Step-by-Step Verified Solution

Equation of trajectory is given as
y = 2x – 9x2     …(A)

Comparing with equation:
y=xtan⁡θ−g2u2cos⁡2θx2y = x\tan \theta - {g \over {2{u^2}{{\cos }^2}\theta }}{x^2}     ...(B)

We get, tan⁡θ=2\tan \theta = 2

∴cos⁡θ=15\therefore \cos \theta = {1 \over {\sqrt 5 }}

Also, g2u2cos⁡2θ=9{g \over {2{u^2}{{\cos }^2}\theta }} = 9

⇒102×9×(15)2=u2;  u2=259\Rightarrow {{10} \over {2 \times 9 \times {{\left( {{1 \over {\sqrt 5 }}} \right)}^2}}} = {u^2};\,\,{u^2} = {{25} \over 9}

⇒u=53m/s\Rightarrow u = {5 \over 3}m/s

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