Question

From the top of a tower of height 40 m, a ball is projected upwards with a speed of 20 $m s^{- 1}$ at an
angle of elevation of 30⁰. The ratio of the total time taken by the ball to hit the ground to
the time taken to come back to the same elevation , is (Take, g = 10 $m s^{- 2}$ )

Moderate

Solution

From $s_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2}, we have$
$- 40 = (20 \sin 30 °) t - 5 t^{2} or t^{2} - 2 t - 8 = 0$
Solving this, we get t = 4 s
$T = \frac{2 u \sin θ}{g} = \frac{2 \times 20 \times \sin 30 °}{10} = 2 s$
$\Rightarrow \frac{t}{T} = 2 : 1$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME