Question:
A stone projected up with a velocity 'u' reaches two points A and B at a distance 'h' with velocities $u/2$ and $u/3$. The maximum height reached by the stone is
A stone projected up with a velocity 'u' reaches two points A and B at a distance 'h' with velocities $u/2$ and $u/3$. The maximum height reached by the stone is
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Max height $H$ is proportional to $u^{2}$. Use the differences in velocity squares to find the relative height $h$.
Updated On: Jul 4, 2026
- $\frac{9h}{5}$
- $\frac{27h}{4}$
- $\frac{36h}{27}$
- $\frac{36h}{5}$
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The Correct Option is D
Solution and Explanation
Step 1: Concept
Use the kinematic equation $v^{2} = u^{2} - 2gh$ and the fact that at maximum height $H$, $H = u^{2}/(2g)$.
Step 2: Meaning
At height $h_{A}$ (point A), $(u/2)^{2} = u^{2} - 2gh_{A}$. At height $h_{B}$ (point B), $(u/3)^{2} = u^{2} - 2gh_{B}$.
Step 3: Analysis
Given the distance between A and B is $h = h_{B} - h_{A}$. From equations: $2gh_{A} = 3u^{2}/4$ and $2gh_{B} = 8u^{2}/9$. Then $2gh = 2g(h_{B}-h_{A}) = u^{2}(8/9 - 3/4) = u^{2}(5/36)$.
Step 4: Conclusion
$u^{2}/2g = H$. Rearranging the previous result gives $H = 36h/5$. Final Answer: (D)
Use the kinematic equation $v^{2} = u^{2} - 2gh$ and the fact that at maximum height $H$, $H = u^{2}/(2g)$.
Step 2: Meaning
At height $h_{A}$ (point A), $(u/2)^{2} = u^{2} - 2gh_{A}$. At height $h_{B}$ (point B), $(u/3)^{2} = u^{2} - 2gh_{B}$.
Step 3: Analysis
Given the distance between A and B is $h = h_{B} - h_{A}$. From equations: $2gh_{A} = 3u^{2}/4$ and $2gh_{B} = 8u^{2}/9$. Then $2gh = 2g(h_{B}-h_{A}) = u^{2}(8/9 - 3/4) = u^{2}(5/36)$.
Step 4: Conclusion
$u^{2}/2g = H$. Rearranging the previous result gives $H = 36h/5$. Final Answer: (D)
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