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Q.Two projectile of same mass and same angle of projection have their maximum kinetic energy in the ratio 4:1 but have different initial velocities, what is the ratio of their horizontal ranges?

(a) 2:1
(b) 4:1
(c) 8:1
(d) 16:1
Manipur CohsemCouncil of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2025MCQ· 1mImportance★★★★★
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Maximum KE of a projectile occurs at launch, so KEmax∝v02KE_{max} \propto v_0^2; range is also R∝v02R \propto v_0^2 (same angle) — so the range ratio equals the KE ratio, 4:1.

For projectile motion, speed is largest at the point of launch and again at the point of landing (both equal to the launch speed v0v_0, ignoring air resistance); it is smallest at the top of the trajectory. So the maximum kinetic energy of the projectile is simply its kinetic energy at launch:

KEmax=12mv02KE_{max} = \tfrac{1}{2}mv_0^2

Given the two projectiles have the same mass and KEmax,1KEmax,2=41\dfrac{KE_{max,1}}{KE_{max,2}} = \dfrac{4}{1}, and mass cancels out:

v012v022=4  ⟹  v01v02=2\frac{v_{01}^2}{v_{02}^2} = 4 \implies \frac{v_{01}}{v_{02}} = 2

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