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September 8th, 2008, 10:05 AM   #1
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A Ball is thrown into the air...

Here's the problem:
Imagine throwing a rock straight up in the air. What is the initial velocity if the rock reaches a max height of 100m?

There is no mention of time, which seems to be the only variable I need to solve the problem. What I have:
A(t)=-10 (to keep it simple)
V(t)=-10t+Vo ----------------------here we have initial velocity. V(t) is 0, but we still can't determine Vo
S(t)=100=-5t^2 + Vot + C---------C is initial height (0m),

When I solve for Vo in S(t) and substitute it in V(t), I get t=sqrt(20), with Vo=44.72, and the book says the answer is 80. Can someone help please? Thanks!
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September 8th, 2008, 01:44 PM   #2
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Re: A Ball is thrown into the air...

If the question really is as you stated, I think you are right.
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September 9th, 2008, 12:52 PM   #3
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Re: A Ball is thrown into the air...

The total energy before throwing it and at it's peak must be the same. Here, I mean the kinetic and potential energy. So, you have:





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January 23rd, 2018, 11:55 AM   #4
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The problem is 100 feet not meters, so you wouldn't use 9.8m but 16ft. You we're doing it right the whole time just with the wrong numbers.
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January 23rd, 2018, 04:04 PM   #5
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Quote:
Originally Posted by hikagi View Post
Here's the problem:
Imagine throwing a rock straight up in the air. What is the initial velocity if the rock reaches a max height of 100m?

There is no mention of time, which seems to be the only variable I need to solve the problem. What I have:
A(t)=-10 (to keep it simple)
V(t)=-10t+Vo ----------------------here we have initial velocity. V(t) is 0, but we still can't determine Vo
S(t)=100=-5t^2 + Vot + C---------C is initial height (0m),

When I solve for Vo in S(t) and substitute it in V(t), I get t=sqrt(20), with Vo=44.72, and the book says the answer is 80. Can someone help please? Thanks!
your work is correct ...

$v_f^2 = v_0^2 - 2g \cdot \Delta y$

at maximum height, $v_f = 0$ ...

$0 = v_0^2 - 2g \cdot \Delta y \implies v_0 = \sqrt{2g \cdot \Delta y} = \sqrt{1960} \approx 44.4 \, m/s$
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