Question

From the top of a tower of height $40m$ , a ball is projected upwards with a speed of $20m{s}^{-1}$ at an angle of elevation of $30^{\circ }$ , then the ratio of the total time taken by the ball to hit the ground to its time of flight upto same horizontal level is :- (Take $g=10mse{c}^{-2}$ )

• A. $2:1$
• B. $3:1$
• C. $3:2$
• D. $4:1$

Answer 1

Answer: (A)

$t=\frac{2usin\theta }{g}=\frac{2\times 20\times sin30^{\circ }}{10}=2s$
Now, we shall calculate the total time taken by the ball to hit the ground.
Using, $s=ut+\frac{1}{2}g{t}^2,$ we get
$40=-10t^{{}^{\prime }}+\frac{1}{2}\times 10\times \left(t^{{}^{\prime }2}\right)$
$\left[\because u=-20sin30^{\circ }=-10m{s}^{-1}\right]$
$\therefore 5\left(t^{{}^{\prime }2}\right)-10t^{{}^{\prime }}-40=0$
Solving, we have, $t^{{}^{\prime }}=4s$
$\therefore \frac{t^{{}^{\prime }}}{t}=\frac{4s}{2s}=\frac{2}{1}$