Question

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

(Take g = 10 m s^-2).

• A. 5 m
• B. 20 m
• C. 15 m
• D. 10 m

Answer 1

Answer: (D)

10 m

Answer 2

Here, $\theta = 30{^\circ},u = 10ms^{- 1}$

$R = 17.3m,g = 10ms^{- 2}$

For horizontal motions, $R = u\cos\theta t$

Or $t = \frac { R } { u \cos \theta } = \frac { 17.3 } { 10 \cos 30 ^ { \circ } } = \frac { 17.3 \times 2 } { 10 \times \sqrt { 3 } } = \frac { 17.3 \times 2 } { 10 \times 1.73 } = 2$

For vertical motions, y $= u\sin\theta t - \frac{1}{2}gt^{2}$

$= 10\sin 30{^\circ} \times 2 - \frac{1}{2} \times 10 \times 2^{2}$

= 10- 20 = -10 m.

Height of tower = 10m.