Question:
A projectile is thrown in upward direction making an angle of 60° with the horizontal direction with a velocity of 150 ms\(^{-1}\). Then, the time after which its inclination with horizontal is 45°, is
A projectile is thrown in upward direction making an angle of 60° with the horizontal direction with a velocity of 150 ms\(^{-1}\). Then, the time after which its inclination with horizontal is 45°, is
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Inclination of projectile at any instant: $\tan\phi = v_y/v_x$. Set $\tan 45° = 1$, so $v_y = v_x$ and solve for time.
Updated On: Apr 20, 2026
- \(15(\sqrt{3}-1)\) s
- \(15(\sqrt{3}+1)\) s
- \(7.5(\sqrt{3}-1)\) s
- \(7.5(\sqrt{3}+1)\) s
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
The horizontal velocity remains constant. The vertical velocity decreases with time. When inclination is 45°, \(v_y = v_x\).
Step 2: Detailed Explanation:
\(v_x = 150\cos60° = 75\) m/s. At inclination 45°: \(v_y = v_x = 75\) m/s. Initial \(v_{y0} = 150\sin60° = 75\sqrt{3}\) m/s. \[ v_y = v_{y0} - gt \Rightarrow 75 = 75\sqrt{3} - 10t \Rightarrow t = \frac{75(\sqrt{3}-1)}{10} = 7.5(\sqrt{3}-1) \text{ s} \]
Step 3: Final Answer:
Time \(= 7.5(\sqrt{3}-1)\) s.
The horizontal velocity remains constant. The vertical velocity decreases with time. When inclination is 45°, \(v_y = v_x\).
Step 2: Detailed Explanation:
\(v_x = 150\cos60° = 75\) m/s. At inclination 45°: \(v_y = v_x = 75\) m/s. Initial \(v_{y0} = 150\sin60° = 75\sqrt{3}\) m/s. \[ v_y = v_{y0} - gt \Rightarrow 75 = 75\sqrt{3} - 10t \Rightarrow t = \frac{75(\sqrt{3}-1)}{10} = 7.5(\sqrt{3}-1) \text{ s} \]
Step 3: Final Answer:
Time \(= 7.5(\sqrt{3}-1)\) s.
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