there is an objest that we drop it streight up.it reaches to the point (h) at the time (t).?

Question

when will it reach to 3/4h from the time we drop it?
1)1/2t 2)3/4t 3)2/5t 3/7t

Answer 1

If it is dropped from a height h and we are asked the time by when it reaches 3/4 h, we can work out this way.

h = ut + 1/2 g.t^2 where h is the height from which it is dropped, and t is the time in seconds. u the initial velocity is zero and g is the acceleration due to gravity.

h = 1/2 gt^2 or t = sqrt. (2h/g)

If we need to find out the time t1 for it to reach a point which is at a height 3/4 h, we can compute the time t1 for 1/4 h.

1/4 h = 0 + 1/2 g. t1^2

1/4 x 1/2g.t^2 = 1/2 gt1^2

Cancelling out 1/2 g from both sides, we get

1/4 . t^2 = t1^2

or t1 = 1/2 t

So answer 1 is correct.

Answer 2

.If we drop it down
the distance S = k T^2 where K is a constant = 0.5g.

¼ S = k ¼ t^2

¼ S = k (t/2)^2

Time taken to fall through ¼ S is t/2

Time to go up through a height of ¾ is 1- t/2 = t/2.

Answer is t/2



However ,we cannot drop it straight up, we have to throw it up.
Question is ambiguous.

If the object goes up through a height H in time t, it will reach a height 3/4 H in time t/2 ..