JEE Main Motion in Plane PYQ

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46. (AIEEE 2003 )

A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle of $30^{\circ}$ with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? $[g = 10 m / s^{2}, \sin 30^{\circ} = \frac{1}{2}, \cos 30^{\circ} = \frac{\sqrt{3}}{2}]$

A. 5.20 m

B. 4.33 m

C. 2.60 m

D. 8.66 m

rrect Option is (D)

AIEEE 2003 Physics - Motion Question 159 English Explanation

From the figure it is clear that maximum horizontal range

$R = \frac{u^{2} \sin 2 θ}{g}$

$= \frac{{(10)}^{2} \sin (2 \times 30^{\circ})}{10}$

$= 5 \sqrt{3}$ = 8.66 m