Question

a ball is projected upwards from the foot of a tower. the ball crosses yhe top of the tower twice after an interval of 6 sec and the ball reaches the ground after 12 sec .calculate the height of the tower...


plz help with full procedure..I will mark the brainliest

Answer 1

Thanks for your question !

     Time taken by ball to go upwards and Then downwards (to the ground) is 12 seconds. Hence, the total time to come down from upwards will be : 12s = 6s + 6s.

    Thus, now, the time taken to reach T and U, and then Again come to T, (passing T two times) is 6 seconds.

     Hence, total time taken to go T upto from the ground = (12 - 6) / 2
                      = 6/2 = 3 Seconds.
    Let the time taken to reach U from ground is : 0 = U - 6 × 10. If we substitute this value in case one above the value of U = 60 m/s,
    Height (H) = 60 × 3 -1/2 × 10 × 3 × 3

    Now, substituting these values in above case 2, the height = 135 m.

     Your answer is - Height of the tower is 135 m.