First slide

Projection Under uniform Acceleration

Question

A ball is thrown from the top of a tower with an initial velocity of 10 ${ms}^{- 1}$ at an angle of $30^{o}$ with the horizontal. If it hits the ground at a distance of 17.3 m from the back of the tower, the height of the tower is (Take g: 10 ${ms}^{- 2}$ )

Moderate

A

5 m
B

20 m
C

15 m
D

10 m

Solution

Here, ${θ= 30}^{o}$ , u = 10 ${ms}^{- 1}$ ,
R = 17.3 m, g = 10 ${ms}^{- 2}$
For horizontal motion, R = $u \cos θt$
$t = \frac{R}{u \cos θ} = \frac{17.3}{10 \cos 30^{0}} = \frac{17.3 \times 2}{10 \times \sqrt{3}} = \frac{17.3 \times 2}{10 \times 1.73} = 2 s$
For vertical motion, h = u sin $θt = \frac{1}{2} {gt}^{2}$
= $10 \sin 30^{0} \times 2 - \frac{1}{2} \times 10 \times 2^{2}$
= 10-20 = -10 m.
Height of tower = 10 m.

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