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Physics-Projectile Motion

NEET Physics Projectile Motion MCQ Question

Type: MCQ-numerical-Medium-Class 11

A projectile is launched with an initial speed vov_o at an angle θo\theta_o with the horizontal. What is the time taken by the projectile to reach its maximum height?

A

vosin⁡θog\frac{v_o \sin \theta_o}{g}

B

2vosin⁡θog\frac{2v_o \sin \theta_o}{g}

C

vocos⁡θog\frac{v_o \cos \theta_o}{g}

D

vog\frac{v_o}{g}

Correct Answer

Option A

Detailed Explanation

The time to reach maximum height (tmt_m) is when vertical velocity vy=0v_y = 0, given by tm=vosin⁡θogt_m = \frac{v_o \sin \theta_o}{g} from Eq. (4.41a) in the NCERT context.

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