A ball is projected with speed 20 m/s at 30° above the horizontal. Taking g = 10 m/s^2, the maximum height reached is

(A) 5 m. Concept. Only the vertical component of velocity decides the height: H = (u_y^2)/(2g). Step 1. Vertical component. u_y = 2030° = 10 m/s. Step 2. Height. H = (10^2)/(2 × 10) = 5 m. Answer. 5 m (Option A).

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Motion in a Plane question for JEE Main, with solution

Q 11-03-001JEE MainTop questionEasy

A ball is projected with speed $20\ \text{m/s}$ at $30^\circ$ above the horizontal. Taking $g = 10\ \text{m/s}^2$, the maximum height reached is

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