Question

19.
A body is projected vertically upwards with a given velocity. The ratio of the times taken to cover the first one
fourth and the last one fourth of the ascent to the highest point is


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Answer 1

Answer:

The velocities a falling object that was projected upwards will be the same as they were at corresponding points of the upward path by conservation of energy (an object projected upwards at V will return to the ground at V)

Let the distance fallen be 1

1/4 = 1/2 g t1^2    t1 = (1 / (2 g))^1/2    time to fall H/4

1/2 = 1/2 g t2^2   t2 = (1/g)^1/2      time to fall to H/2

3/4 = 1/2 g t3^2    t3 = (3 / (2 g))^1/2      time to fall to 3 H / 4

1 = 1/2 g t4^2        t4 = (2 / g)^1/2

Now we want    (t4 - t3) / t1

[2^1/2 - (3/2)^1/2] / (1/2)^1/2 = 2 - 3^1/2 = .268