Question

From the top of a tower of height $40m$ , a ball is projected upwards with a speed of $20ms^{- 1}$ 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 $g=10msec^{- 2}$ )

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

Answer 1

Answer: (A)

$t=\frac{2 u s i n \theta }{g}=\frac{2 \times 20 \times s i n 30 ^\circ }{10}=2s$
Now, we shall calculate the total time taken by the ball to hit the ground.
Using, $s=ut+\frac{1}{2}gt^{2},$ we get
$40=-10t^{'}+\frac{1}{2}\times 10\times \left(t^{' 2}\right)$
$\left[\because u = - 20 sin 30 ^\circ = - 10 m s^{- 1}\right]$
$\therefore 5\left(t^{' 2}\right)-10t^{'}-40=0$
Solving, we have, $t^{'}=4s$
$\therefore \frac{t^{'}}{t}=\frac{4 s}{2 s}=\frac{2}{1}$