Question:
Two seconds after projection, a projectile is travelling in a direction inclined at \(30^\circ\) to the horizontal. After one more sec, it is travelling horizontally. Find its velocity magnitude and direction.
Two seconds after projection, a projectile is travelling in a direction inclined at \(30^\circ\) to the horizontal. After one more sec, it is travelling horizontally. Find its velocity magnitude and direction.
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Horizontal motion $\Rightarrow$ vertical velocity = 0 $\Rightarrow$ use it to find initial components.
Updated On: Apr 23, 2026
- \(2\sqrt{20}\,m/s,\ 60^\circ\)
- \(20\sqrt{3}\,m/s,\ 60^\circ\)
- \(40\,m/s,\ 30^\circ\)
- \(40\sqrt{6}\,m/s,\ 30^\circ\)
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The Correct Option is B
Solution and Explanation
Concept:
At highest point, vertical velocity = 0.
Step 1: At \(t=3s\), motion is horizontal
\[ u\sin\theta = g \cdot 3 \]
Step 2: At \(t=2s\), direction \(30^\circ\)
\[ \tan 30^\circ = \frac{u\sin\theta - 2g}{u\cos\theta} \]
Step 3: Solve
\[ u\sin\theta = 3g,\quad u\cos\theta = \sqrt{3}g \] \[ u = \sqrt{(3g)^2 + (g\sqrt{3})^2} = 20\sqrt{3} \] \[ \theta = 60^\circ \]
Step 1: At \(t=3s\), motion is horizontal
\[ u\sin\theta = g \cdot 3 \]
Step 2: At \(t=2s\), direction \(30^\circ\)
\[ \tan 30^\circ = \frac{u\sin\theta - 2g}{u\cos\theta} \]
Step 3: Solve
\[ u\sin\theta = 3g,\quad u\cos\theta = \sqrt{3}g \] \[ u = \sqrt{(3g)^2 + (g\sqrt{3})^2} = 20\sqrt{3} \] \[ \theta = 60^\circ \]
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