Question

Two projectiles of same mass and with same velocity are thrown at an angle \(60º\) & \(30º\) with the horizontal, then which quantity will remain same:

• A. Time of flight
• B. Horizontal range of projectile
• C. Max height acquired
• D. All of them

Answer 1

The Correct Option is B

To determine which quantity remains the same when two projectiles are thrown at angles \(60^\circ\) and \(30^\circ\) with the horizontal, we must understand the physics of projectile motion.

The key quantities in projectile motion are:

• Time of Flight (T): The total time the projectile is in the air.
• Horizontal Range (R): The horizontal distance covered by the projectile.
• Maximum Height (H): The highest point reached by the projectile.

Given that both projectiles have the same mass and initial velocity, we'll analyze which quantity remains unchanged between the two angles.

Time of Flight:

The time of flight for a projectile launched at an angle \(\theta\) is given by:

T = \frac{2u \sin \theta}{g}

where \(u\) is the initial velocity and \(g\) is the acceleration due to gravity.

For \(\theta = 30^\circ\) and \(\theta = 60^\circ\):

• T_{30} = \frac{2u \sin 30^\circ}{g} = \frac{u}{g}
• T_{60} = \frac{2u \sin 60^\circ}{g} = \frac{\sqrt{3}u}{g}

The times of flight are different.

Horizontal Range:

The horizontal range is given by:

R = \frac{u^2 \sin 2\theta}{g}

For \(\theta = 30^\circ\):

R_{30} = \frac{u^2 \sin 60^\circ}{g}

For \(\theta = 60^\circ\):

R_{60} = \frac{u^2 \sin 120^\circ}{g} = \frac{u^2 \sin 60^\circ}{g}

The horizontal ranges are the same because \(\sin 60^\circ = \sin 120^\circ\).

Maximum Height:

The maximum height is given by:

H = \frac{u^2 \sin^2 \theta}{2g}

For \(\theta = 30^\circ\) and \(\theta = 60^\circ\):

• H_{30} = \frac{u^2 (\sin 30^\circ)^2}{2g} = \frac{u^2}{8g}
• H_{60} = \frac{u^2 (\sin 60^\circ)^2}{2g} = \frac{3u^2}{8g}

The maximum heights are different.

Therefore, the quantity that remains the same for both angles is the Horizontal Range of the projectile.