Back to Practice
Physics-Projectile Motion

NEET Physics Projectile Motion MCQ Question

Type: MCQ-diagram based-Medium-Class 11

Imagine a projectile launched from the origin at an angle θ0\theta_0 with initial velocity v0v_0. Which of the following equations correctly represents the time tmt_m taken to reach maximum height?

Question diagram
A

tm=v0sin⁡θ0gt_m = \frac{v_0 \sin \theta_0}{g}

B

tm=2v0cos⁡θ0gt_m = \frac{2v_0 \cos \theta_0}{g}

C

tm=v0cos⁡θ0gt_m = \frac{v_0 \cos \theta_0}{g}

D

tm=2v0sin⁡θ0gt_m = \frac{2v_0 \sin \theta_0}{g}

Correct Answer

Option A

Detailed Explanation

The time to reach maximum height tmt_m is given by tm=v0sin⁡θ0gt_m = \frac{v_0 \sin \theta_0}{g} because at maximum height, the vertical velocity component becomes zero.

Found an issue with this question?

Related Questions

Explore More Questions

All Physics QuestionsMotion in a Plane Questions