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JEE Main 2013 (22 Apr, Shift 1) · Projectile Basics & Trajectory

Projectile Basics & Trajectory question from JEE Main 2013 (22 Apr, Shift 1)

Q8JEE Main 2013 · 22 APR · SHIFT 1Difficult

Projectile Basics & Trajectory

A ball projected from ground at an angle of 45° just clears a wall in front. If point of projection is 4 m from the foot of wall and ball strikes the ground at a distance of 6 m on the other side of the wall, the height of the wall is:

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Answer: (b)

Range $R = 4 + 6 = 10$ m; trajectory $y = x\tan\theta\left(1 - \frac xR\right)$
At the wall $x = 4$ m, $\theta = 45^\circ$ : $y = 4\times1\times\left(1 - \frac{4}{10}\right)$
$y = 2.4$ m
So the answer is (b) 2.4 m.

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