Question

An object is projected with an angle of \( 60^{\circ} \) with horizontal with a velocity V. During the path, when it makes \( 30^{\circ} \) with horizontal, its velocity becomes 10 \( ms^{-1} \), then V is:

• A. \( 10\sqrt{3} \, ms^{-1} \)
• B. \( \sqrt{30} \, ms^{-1} \)
• C. \( 3\sqrt{10} \, ms^{-1} \)
• D. \( \sqrt{3} \, ms^{-1} \)

Hint:

The key to projectile motion problems is treating the horizontal component as uniform motion. Since there is no acceleration along the x-axis, \( v_x = \text{constant} \) will quickly unlock the relation between speeds at different angles.

Answer 1

The Correct Option is A

Step 1: Pick the key idea of projectile motion.
In projectile motion gravity pulls only straight down. So there is no force sideways. This means the horizontal part of the velocity never changes during the whole flight. We will use this single fact.
Step 2: Write the horizontal speed at launch.
The object is launched with speed $V$ at $60^{\circ}$ above the ground. The horizontal part is: \[ V_x = V\cos 60^{\circ} \]
Step 3: Write the horizontal speed at the later point.
Later, the speed is $10\ ms^{-1}$ and the direction is $30^{\circ}$ above the ground. Its horizontal part is: \[ v_x = 10\cos 30^{\circ} \]
Step 4: Make them equal.
Since horizontal speed stays constant, the two must be equal: \[ V\cos 60^{\circ} = 10\cos 30^{\circ} \]
Step 5: Put in the standard values.
Use $\cos 60^{\circ} = \tfrac{1}{2}$ and $\cos 30^{\circ} = \tfrac{\sqrt{3}}{2}$: \[ V\left(\frac{1}{2}\right) = 10\left(\frac{\sqrt{3}}{2}\right) \]
Step 6: Solve for V.
The factor $\tfrac{1}{2}$ cancels from both sides, giving: \[ \boxed{V = 10\sqrt{3}\ ms^{-1}} \]