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Physics Question on Motion in a plane

From the top of a tower of height 40 m, a ball is projected upwards with a speed of 20 m s−1{20 \,m \, s^{-1}} at an angle of elevation of 30?. Then the ratio of the total time taken by the ball to hit the ground to its time to same height is Take g=10 m s−2{g = 10 \, m \, s^{-2}}

A

2:01

B

3:01

C

3:02

D

4:01

Answer

2:01

Explanation

Solution

If tt is the total time taken, then
40=−20 sin⁡ 30∘ t+12×10×t240 = - 20 \, \sin \, 30^{\circ}\, t + \frac{1}{2} \times 10 \times t^2
40=−20sin⁡ 30∘t+5t240 = -20 \sin \, 30^\circ t + 5t^2
or 40=−10t+5t240 = - 10 t + 5t^2
or 5t2−10t−40=05 t^2 - 10t - 40 = 0
or t2−2t−8=0t^2 - 2t - 8 = 0
or t2−4t+2t−8=0t^2 - 4t + 2t - 8 = 0
or t(t−4)+2(t−4)=0t(t - 4) + 2(t - 4) = 0
or (t+2)(t−4)=0(t + 2)(t - 4) = 0
⇒ t=4 s\Rightarrow \, t = 4 \, s [Negative time is not allowed]
Time of flight to height of projection
T=2u sin⁡ θgT= \frac{2u \, \sin \, \theta}{g}
=2×20 sin⁡ 30∘10s=2s= \frac{2 \times 20 \, \sin \, 30^\circ}{10} s = 2s
∴     tT=42=21\therefore \:\:\:\:\: \frac{t}{T} = \frac{4}{2} = \frac{2}{1}