From top of a tower of height 40,m , a ball is projected upwards with a speed of 20,m / s at an angl

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3-2.Motion in Plane

hard

From the top of a tower of height $40\,m$, a ball is projected upwards with a speed of $20\,m / s$ at an angle of elevation of $30^{\circ}$. 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 $g=10\,m / s ^2$ )

A

$2: 1$

B

$3: 1$

C

$3: 2$

D

$1.5: 1$

Solution

(a)

From $s_y=u_y t+\frac{1}{2} a_y t^2$ we have

$-40=\left(20 \sin 30^{\circ}\right) t-5 t^2$

or $\quad t^2-2 t-8=0$

Solving this, we get $t=4\,s$

$T =\frac{2 u \sin \theta}{g}$

$=\frac{2 \times 20 \times \sin 30^{\circ}}{10}=2\,s$

$\frac{t}{T} =2$

Standard 11

Physics

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(AIEEE-2002)