Velocity from height and time (stone thrown from building)

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skatru

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I don't know how to start. Just a little help to get me started on this problem please.

A stone is thrown down a skyscraper. It passes a window after 2 seconds. The window is at a height of 450 ft. It hits the ground after 5 seconds (from the time it was thrown). Find the velocity the stone was thrown down at and the height it was thrown from. (Neglect air resistance.)

That's the info I'm given and I don't know how to start.

Thanks
 
skatru said:
A stone is thrown down a skyscraper. It passes a window after 2 seconds. The window is at a height of 450 ft. It hits the ground after 5 seconds (from the time it was thrown). Find the velocity the stone was thrown down at and the height it was thrown from.
Click to expand...
You know that velocity is the integral (or antiderivative) of acceleration, and that the only acceleration on the object (once released) is due to gravity. You need to find the initital velocity, v[sub:1xjsrdd3]0[/sub:1xjsrdd3] = v(0), given that a(t) = -32 ft/sec[sup:1xjsrdd3]2[/sup:1xjsrdd3].

You know that position is the integral (or antiderivative) of velocity, and that the original position, s(0) = s[sub:1xjsrdd3]0[/sub:1xjsrdd3], is also the height of the building. Assuming "after five seconds" means "at the end of five-seconds duration", then s(5) = 0. Also, you are given that s(2) = 450.

How far have you gotten with this information? Please be complete. Thank you! :D

Eliz.
 
I'm pretty sure I understand how to get height if I could figure out velocity.

The integral of acceleration would be v(t) = -32t + C. So if V(0) -32(0) + C then C = O but I know that the initial velocity is supposed to be -38 ft/sec. I'm pretty sure that I'm missing something here.
 
(I will assume that you mean V and v to actually indicate the same thing. This is contrary to standard mathematical practice, so if my assumption is in error, kindly forgive my confusion and reply with correction.)

skatru said:
The integral of acceleration would be v(t) = -32t + C. So if V(0) -32(0) + C then C = O....
Click to expand...
How did you arrive at C = 0? This would require that, for v(t) = -32t + C, we have v(0) = 0 + C = 0. But clearly we do not. Instead, we know that the object was "thrown" (not "dropped"), and that therefore the initial velocity was some non-zero value.

Please clarify your steps and reasoning for the result you obtained. Thank you! :D

Eliz.
 
Yeah, I meant V to be v. Sorry about not being consistent.

That is what I did with the equation. v(0) = 0 + C = 0. Is v(t)= -32t+C right? Since v(0) is supposed to equal the initial velocity. I thought if I put t=0 in v(t)=-32t+C then I get C = 0 which I know isn't right. So have I explained where I'm struck in the equation in a decent enough way.
 
A stone is thrown down a skyscraper. It passes a window after 2 seconds. The window is at a height of 450 ft. It hits the ground after 5 seconds (from the time it was thrown). Find the velocity the stone was thrown down at and the height it was thrown from. (Neglect air resistance.)

Vo = the initial velocity
V1 = the velocity at the window
V2 = the ground impact velocity
From h = Vot + gt^2/2 and
The time from the window to the ground is 3 seconds
450 = V1(3)+16(3)^2 from which Vo = 102fps.
From V1 = Vo+gt, 102 = Vo+32(2)from which Vo = 38fps.
As for the height from the roof to the window:
Again, from h = Vot + 16t^2, h = 38(2) + 16(2)^2 = 140ft.
Therefore, the height of thje building is 140 + 450 = 590ft.

Thus, the stone was thrown from a height of 590 ft. with an initial downward velocity of 38 fps.
 
Distance Formula: S(t) = -16t^2+V_o(t) + S_o.

S = distance from ground at time t, V_o = initial velocity, S_o = initial distance from ground. Let x+450 = S_o.

Hence S(t) = -16t^2 + V_o(t) + 450 + x. S(2) = 450 = -64 + 2V_o + 450 + x. x = 64 - 2V_o.

Ergo S(t) = -16t^2 +V_o(t-2) + 514. S(5) = 0 = -400 + 3V_o + 514. 0 = 3V_o = -114, V_o = -38, x = 64 - 2 (-38) = 140

Therefore S(t) = -16t^2 -38t + (140+450), S(t) = -16t^2 -38t +590.

The stone is thrown at an initial velocity of -38ft/sec from a height of 590ft.
 
Thank You. I see what I wasn't doing. I needed to get the integral of v(t) to find everything. Thanks!
 
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