Question:
A stone is thrown at an angle to the horizontal reaches a maximum height H. Then the time of flight of stone will be
A stone is thrown at an angle to the horizontal reaches a maximum height H. Then the time of flight of stone will be
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A stone is thrown at an angle to the horizontal reaches a maximum height H. Then the time of flight of stone will be
Updated On: Apr 15, 2026
- $\sqrt{\frac{2H}{g}}$
- $2 \sqrt{\frac{2H}{g}}$
- $\frac{2\sqrt{2H}\sin \theta}{g}$
- $\frac{\sqrt{2H}\sin \theta}{g}$
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The Correct Option is B
Solution and Explanation
Step 1: Formulas
Max height $H = \frac{u^{2}\sin^{2}\theta}{2g}$ and Time of flight $T = \frac{2u\sin\theta}{g}$.
Step 2: Relation
Squaring $T$ gives $T^{2} = \frac{4u^{2}\sin^{2}\theta}{g^{2}}$. Dividing $T^{2}$ by $H$ gives $\frac{T^{2}}{H} = \frac{8}{g}$.
Step 3: Calculation
$T^{2} = \frac{8H}{g} \Rightarrow T = \sqrt{\frac{8H}{g}} = 2\sqrt{\frac{2H}{g}}$.
Final Answer: (b)
Max height $H = \frac{u^{2}\sin^{2}\theta}{2g}$ and Time of flight $T = \frac{2u\sin\theta}{g}$.
Step 2: Relation
Squaring $T$ gives $T^{2} = \frac{4u^{2}\sin^{2}\theta}{g^{2}}$. Dividing $T^{2}$ by $H$ gives $\frac{T^{2}}{H} = \frac{8}{g}$.
Step 3: Calculation
$T^{2} = \frac{8H}{g} \Rightarrow T = \sqrt{\frac{8H}{g}} = 2\sqrt{\frac{2H}{g}}$.
Final Answer: (b)
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