First slide

Projection Under uniform Acceleration

Question

From the top of a tower of height 40 m a ball is projected upwards with a speed of 20 m/sec at an angle of elevation of 30^o. Then the ratio of the total time taken by the ball to hit the ground to its time of flight (time taken to come back to the same elevation) is (take g = 10 m/sec²)

Moderate

A

2 : 1
B

3 : 1
C

3 : 2
D

4 : 1

Solution

$t = \frac{2 usin θ}{g} = \frac{2 \times 20 \sin 30}{10} = 2 s$
Now we shall calculate the total time taken by the ball to hit the ground. Using $s = ut + \frac{1}{2} {gt}^{2}$ , we have
$\begin{array}{l} 40 = - 10 t^{'} + \frac{1}{2} \times 10 \times {(t^{'})}^{2} \\ ∵ 4 = - 20 \sin 30^{\circ} = - 10 m / s \\ or 5 {(t^{'})}^{2} - 10 t^{'} - 40 = 0 \\ or {(t^{'})}^{2} - 2 t^{'} - 8 = 0 or t^{'} = 4 s \\ ∴ \frac{t^{'}}{t} = \frac{4 s}{2 s} = \frac{2}{1} \end{array}$

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