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JEE Main 2022 · Physics · Kinematics and Laws of Motion · easy

A ball is thrown from the top of a vertical cliff 20 m high with speed 25 m s^-1 at an angle of 37^° above the horizontal (sin 37^°=0.6, g=10 m…

Single correct option

A ball is thrown from the top of a vertical cliff 20 m20\ \mathrm{m} high with speed 25 m s−125\ \mathrm{m\,s^{-1}} at an angle of 37∘37^\circ above the horizontal (sin⁡37∘=0.6\sin37^\circ=0.6, g=10 m s−2g=10\ \mathrm{m\,s^{-2}}). It lands on level ground at the foot of the cliff. The horizontal distance of the landing point from the foot of the cliff is
  1. A. 60 m60\ \mathrm{m}
  2. B. 80 m80\ \mathrm{m}
  3. C. 100 m100\ \mathrm{m}
  4. D. 120 m120\ \mathrm{m}
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Answer: (B) 80 m80\ \mathrm{m}

Components: ux=25cos⁡37∘=20 m s−1u_x=25\cos37^\circ=20\ \mathrm{m\,s^{-1}} and uy=25sin⁡37∘=15 m s−1u_y=25\sin37^\circ=15\ \mathrm{m\,s^{-1}}. The vertical displacement on landing is −20 m-20\ \mathrm{m}: −20=15t−5t2-20=15t-5t^{2}, i.e. t2−3t−4=0t^{2}-3t-4=0, so t=4 st=4\ \mathrm{s} (the negative root is rejected). Horizontal distance =uxt=20×4=80 m=u_xt=20\times4=80\ \mathrm{m}. The level-ground range formula cannot be used because the launch and landing heights differ.
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