If a stone is dropped from a balloon rising at the rate of 20 fps reaches the ground in 10 seconds, find the initial height and final velocity.

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embizze's profile pic

Posted on

The equation for a free falling body (no acceleration except gravity -- thus ignoring air resistance, etc...) is

`h(t)=-16t^2+v_0t+h_0` where

h -- the height in feet at time t

t -- the time in seconds

`v_0` -- the initial velocity in feet per second.. (Positive if away from the Earth, and negative if towards the Earth.)

`h_0` -- the initial height in feet.

The velocity function is the derivative of the displacement function so with v as the velocity in feet per second at time t we get:


To find the initial height, we use the given information that at time t=10 the height is 0 and the initial velocity is 10fps (again positive since it is directed away from the ground):



To find the final velocity we substitute t=10 into the velocity formula:

`v_(10)=-32(10)+10` or v=-310fps.


The initial height was 1500 ft and the final velocity was -310 fps.


llltkl's profile pic

Posted on

Well, that also solves the problem. The initial velocity of the balloon (and hence the initial vertical velocity of the stone), is 20 ft/s (Given), but you have taken it as 10 ft/s whlie calculating `h_0` .

You would get the same results as mine through this method too, once you take accurate values of g (I have taken 32.2 ft/s^2), and `v_0` for calculation.

Top Answer

llltkl's profile pic

Posted on

When the stone is dropped from the rising balloon, at first it will have two forces acting on it in opposite directions: one upward due to initial upward velocity and the other downward acceleration due to gravity. As a consequence, the stone will rise a bit and then start falling to the ground freely.
While on the rise,

u=20 ft/s
v = 0(since at the topmost point its velocity will become zero)
g=-32.2 ft/s^2(since g is downwards)
we get time to rise up, t=0.621118 s

Height attained in this time, in addition to the initial height, can be obtained by applying

`s = ut-1/2 gt^2`

`=20*0.621118-1/2 *32.2*(0.62118)^2 = 6.21118 ft.`

So, total height attained by the stone before coming to a standstill and then beginning to fall

= (h+ 6.21118) ft., where h is the initial height.

While experiencing the free fall, let t be the time taken to reach the ground from the topmost point of its flight.


`(h+ 6.21118) = 0*t+1/2 g t^2` (here, g will be positive)

`rArr t^2 = (h+ 6.21118)/16.1`

`rArr t = sqrt((h+ 6.21118)/16.1)`

Total time of flight of the stone

`= sqrt((h+ 6.21118)/16.1)+0.621118`

By condition, `sqrt((h+ 6.21118)/16.1)+0.621118 = 10`

Upon solving, we get h = 1410 ft.

Final velocity before touching the ground, at the end f the free fall  can be obtained by applying

 `v^2 = u^2 +2gs`

Plugging in the values,

`v^2 = 0 + 2*32.3*1416.21118`

`rArr v = 302` ft./s

Therefore, the initial height of the stone, from where it was dropped, was 1410 ft. and its final velocity  before touching the ground was 302 ft./s.


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tobydator's profile pic

Posted on

Thanks a lot:))

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