⚛️ Physics · Motion in a Plane · NEET & JEE

The maximum height reached by a projectile launched at speed u and angle θ is:

Answer: u²sin²θ / 2g.

A u²sin²θ / 2g
B u²sin2θ / g
C u sinθ / g
D u² / 2g

Correct answer: A. u²sin²θ / 2g

Explanation: Maximum height H = u²sin²θ / 2g.

Projectile trajectories launched at 15, 30, 45, 60, 75 and 90 degrees at the same speed, each annotated with range R, maximum height H and time of flight T, with coloured dots marking equal time intervals

Projectiles at equal speed: range is maximum at 45°, and complementary angles (30° and 60°) share the same range. Image: Cmglee, CC BY-SA 3.0, via Wikimedia Commons.

Concept context

Vectors, projectile motion, uniform circular motion, and relative velocity - the foundation of 2D kinematics for NEET and JEE.

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The maximum height reached by a projectile launched at speed u and angle θ is:

Concept context

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