Physics-Projectile Motion

NEET Physics Projectile Motion MCQ Question

Type: MCQ-diagram based-Medium-Class 11

Consider a diagram showing the trajectory of a projectile launched from the origin with an initial velocity v0v_0 at an angle θ0\theta_0 with the horizontal. If vyv_y is the vertical component of velocity at a certain time tt, what is the time tmt_m at which vyv_y becomes zero?

Question diagram
A

tm=v0sinθ0gt_m = \frac{v_0 \sin \theta_0}{g}

B

tm=v0cosθ0gt_m = \frac{v_0 \cos \theta_0}{g}

C

tm=2v0sinθ0gt_m = \frac{2v_0 \sin \theta_0}{g}

D

tm=v0tanθ0gt_m = \frac{v_0 \tan \theta_0}{g}

Correct Answer

Option A

Detailed Explanation

The time to reach maximum height where the vertical component of velocity is zero is given by tm=v0sinθ0gt_m = \frac{v_0 \sin \theta_0}{g}, based on the equation vy=v0sinθ0gtm=0v_y = v_0 \sin \theta_0 - g t_m = 0.

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