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.
3 Answers
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:
`v(t)=-32t+v_0`
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):
`0=-16(10)^2+10(10)+h_0`
`=>h_0=1500"ft"`
To find the final velocity we substitute t=10 into the velocity formula:
`v_(10)=-32(10)+10` or v=-310fps.
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The initial height was 1500 ft and the final velocity was -310 fps.
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llltkl | College Teacher | (Level 3) Valedictorian
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.
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)
applying
v=u-gt
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.
Then,
`(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 | Student, College Freshman | (Level 1) Honors
Thanks a lot:))