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 ${ms}^{- l}$ at an angle of elevation of $30^{o}$ . Then the ratio of the total time taken by the ball to hit the ground to the time taken to ball come at same level as top of tower.

Moderate

A

2 : 1
B

3 : 1
C

3 : 2
D

4 : 1

Solution

If t is the total time taken, then
$40 = - 20 \sin 30^{0} t + \frac{1}{2} \times 10 \times t^{2}$
$or 40 = - 10 t + 5 t^{2}$
$or 5 t^{2} - 10 t - 40 = 0$
or $t^{2} - 2 t - 8 = 0$
or $t^{2} - 4 t + 2 t - 8 = 0$
or $t (t - 4) + 2 (t - 4) = 0$
or (t+2)(t-4) = 0
t = 4 s [Negative time is not allowed]
$T = \frac{2 v \sin θ}{g} = \frac{2 \times 20 \sin 30^{0}}{10} s = 2 s$
$∴ \frac{t}{T} = \frac{4}{2} = \frac{2}{1}$

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