From the top of a tower of height $$40$$ m a ball is projected upwards with a speed of $$20$$ $$m/$$$$sec$$ at an angle of elevation of $$30o$$. Then 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/sec2)

Question

Infinity Learn · Accepted Answer

$$t=2usin$$⁡θ$$g=2$$×$$20sin$$⁡$$3010=2sNow$$ we shall calculate the total time taken by the ball to hit the ground. Using $$s=ut+12gt2$$, we have $$40=$$−$$10t$$′$$+12$$×$$10$$×t′$$2$$∵ $$4=$$−$$20sin$$⁡$$30$$∘$$=$$−$$10m/s$$ or  $$5t$$′$$2$$−$$10t$$′−$$40=0$$ or  t′$$2$$−$$2t$$′−$$8=0$$  or  t′$$=4s$$∴ t′$$t=4s2s=21$$

From the top of a tower of height 40 m a ball is projected upwards with a speed of 20 m/sec at an angle of elevation of 30^o. Then 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/sec²)

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From the top of a tower of height 40 m a ball is projected upwards with a speed of 20 m/sec at an angle of elevation of 30o. Then 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/sec2)

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From the top of a tower of height 40 m a ball is projected upwards with a speed of 20 m/sec at an angle of elevation of 30^o. Then 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/sec²)