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Standard XII

Physics

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Question

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 $30^{0}$ . 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 :

2 : 1

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1 : 2

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3 : 2

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Solution

The correct option is A $2 : 1$
Time of flight,
$T = \frac{2 v sin θ}{g} = \frac{2 \times 20 \times sin 30 °}{10} = 2 s e c$
Total time taken by the ball to hit the ground is
$t = \frac{d i s t a n c e}{v e l o c i t y} = \frac{40}{20 sin 30 °} = \frac{40}{20 \times \frac{1}{2}} = 4 s e c$
So, the ratio is $4 : 2 = 2 : 1$

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From the top of a tower of height 40m, a ball is projected upwards with a speed of 20m/s at an angle of elevation of 300. 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 :

The correct option is A 2:1Time of flight,T=2vsinθg=2×20×sin30°10=2secTotal time taken by the ball to hit the ground is t=distancevelocity=4020sin30°=4020×12=4secSo, the ratio is 4:2=2:1

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 $30^{0}$ . 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 :