From the top of a tower of height $$40$$ m, a ball is projected upwards with a speed of $$20$$ $$ms-1$$ at anangle of elevation of $$300$$. The ratio of the total time taken by the ball to hit the ground to the time taken to come back to the same elevation , is (Take$$, g $$=$$ $$10$$ $$ms-2)

Question

From the top of a tower of height $$40$$ m, a ball is projected upwards with a speed of $$20$$ $$ms-1$$ at anangle of elevation of $$300$$. The ratio of the total time taken by the ball to hit the ground to the time taken to come back to the same elevation , is (Take$$, g $$=$$ $$10$$ $$ms-2)

Infinity Learn · Accepted Answer

From $$sy=uyt+12ayt2$$, we have $$-40=20sin30$$°$$t-5t2$$ or $$t2-2t-8=0Solving$$ this, we get t $$=$$ $$4$$ $$sT=2usin$$θ$$g=2$$×$$20$$×$$sin30$$°$$10=2$$ s⇒  $$tT=2$$:$$1$$

From the top of a tower of height 40 m, a ball is projected upwards with a speed of 20 $m s^{- 1}$ at an
angle of elevation of 30⁰. The ratio of the total time taken by the ball to hit the ground to
the time taken to come back to the same elevation , is (Take, g = 10 $m s^{- 2}$ )

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From the top of a tower of height 40 m, a ball is projected upwards with a speed of 20 ms-1 at anangle of elevation of 300. The ratio of the total time taken by the ball to hit the ground to the time taken to come back to the same elevation , is (Take, g = 10 ms-2)

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Ready to Test Your Skills?

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From the top of a tower of height 40 m, a ball is projected upwards with a speed of 20 $m s^{- 1}$ at an
angle of elevation of 30⁰. The ratio of the total time taken by the ball to hit the ground to
the time taken to come back to the same elevation , is (Take, g = 10 $m s^{- 2}$ )