A body falls freely from the top of a tower. It covers 36% of the height?

Question

... in the last second before striking the ground level. What would be the height of the tower . Given that g=10 m/s square.


please help thanks

Answer 1

Let H be the height of the building and t be the time to fall through this height H.

Then H = ½ g t^2 = 5 t^2. ----------------- (1)

In (t-1) second the distance traveled is 1 - 0.36 H = 0.64 H.

0.64 H = 5 (t-1)^2.---------------------------(2)

(2) / (1) gives 0.64 = (t-1)^2/ t^2

Or 0.8 = (t-1) / t

Solving we get t= 5 second.

Using (1) we get H = 125 m.

Answer 2

I don't think you have enough information to solve. If the tower has height h and the object falls for time t, at time t - 1 the object has a velocity V equal to g*(t - 1) or 10t - 10. The distance that the object falls during the last second, a time of duration 1, is V*1 + 0.5*g*1^2 = 10t - 10 + 5 = 10t - 5. If you set 10t - 5 equal to 0.36h, you still have two unknowns, t and h.

Answer 3

You do have enough information. Let h be the height of the tower.

For the last second of motion use:

s = vt + 1/2 gt^2, with t = 1 of course

-> 0.36h = v + 1/2g = v + 5

Now v is the speed it has reached just before its last second of motion. To find this use:

v^2 = u^2 + 2gs, with the body starting at rest at the top (u=0)

-> v = sqrt(2gs) = sqrt(2g . 0.64h) = sqrt(12.8h)

Substitute:

-> 0.36h = sqrt(12.8h) + 5

You can turn this into a quadratic and solve it.

Answer 4

Answer He was a dwarf