Question:
The time taken by a body projected vertically upwards from the ground to reach 75% of the maximum height it can reach is 2 s. The ratio of the kinetic energy of the body at a time $t=2$ s and the potential energy of the body at a time $t=3$ is (Acceleration due to gravity $=10~ms^{-2}$)
The time taken by a body projected vertically upwards from the ground to reach 75% of the maximum height it can reach is 2 s. The ratio of the kinetic energy of the body at a time $t=2$ s and the potential energy of the body at a time $t=3$ is (Acceleration due to gravity $=10~ms^{-2}$)
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When both KE and PE are asked, eliminate mass first to simplify ratios.
Updated On: Jun 22, 2026
- 2 : 15
- 4 : 15
- 2 : 5
- 4 : 5 \bigskip
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The Correct Option is B
Solution and Explanation
Concept:
For vertical projectile motion, kinematic equations together with energy conservation are used. Height, velocity, kinetic energy (KE) and potential energy (PE) are interrelated through $v^2 = u^2 - 2gy$ and total mechanical energy conservation.
Step 1: Let initial velocity be $u$.
Maximum height is: \[ H = \frac{u^2}{2g} \]
Step 2: Use condition for 75% height at $t=2$ s.
Given body reaches $0.75H$ in 2 s: \[ s = ut - \frac{1}{2}gt^2 \] At $t=2$: \[ 0.75 \cdot \frac{u^2}{2g} = 2u - \frac{1}{2}g(4) \] Substitute $g=10$: \[ 0.75 \cdot \frac{u^2}{20} = 2u - 20 \] Multiply by 20: \[ 0.75u^2 = 40u - 400 \] \[ 3u^2 = 160u - 1600 \] \[ 3u^2 - 160u + 1600 = 0 \] Solving gives: \[ u = 40~m/s \]
Step 3: Velocity at $t=2$ s.
\[ v = u - gt = 40 - 20 = 20~m/s \] KE at $t=2$: \[ KE_2 = \frac{1}{2}m(20^2) = 200m \]
Step 4: Position at $t=3$ s.
\[ s = 40(3) - 5(9) = 120 - 45 = 75 \] PE at $t=3$: \[ PE_3 = mg(75) \]
Step 5: Ratio.
\[ KE_2 : PE_3 = 200m : 750m = 4 : 15 \]
Step 1: Let initial velocity be $u$.
Maximum height is: \[ H = \frac{u^2}{2g} \]
Step 2: Use condition for 75% height at $t=2$ s.
Given body reaches $0.75H$ in 2 s: \[ s = ut - \frac{1}{2}gt^2 \] At $t=2$: \[ 0.75 \cdot \frac{u^2}{2g} = 2u - \frac{1}{2}g(4) \] Substitute $g=10$: \[ 0.75 \cdot \frac{u^2}{20} = 2u - 20 \] Multiply by 20: \[ 0.75u^2 = 40u - 400 \] \[ 3u^2 = 160u - 1600 \] \[ 3u^2 - 160u + 1600 = 0 \] Solving gives: \[ u = 40~m/s \]
Step 3: Velocity at $t=2$ s.
\[ v = u - gt = 40 - 20 = 20~m/s \] KE at $t=2$: \[ KE_2 = \frac{1}{2}m(20^2) = 200m \]
Step 4: Position at $t=3$ s.
\[ s = 40(3) - 5(9) = 120 - 45 = 75 \] PE at $t=3$: \[ PE_3 = mg(75) \]
Step 5: Ratio.
\[ KE_2 : PE_3 = 200m : 750m = 4 : 15 \]
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