September 8th, 2008, 10:05 AM  #1 
Newbie Joined: Oct 2007 Posts: 7 Thanks: 0  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 + CC 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! 
September 8th, 2008, 01:44 PM  #2 
Senior Member Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0  Re: A Ball is thrown into the air...
If the question really is as you stated, I think you are right.

September 9th, 2008, 12:52 PM  #3 
Senior Member Joined: May 2007 Posts: 402 Thanks: 0  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: 
January 23rd, 2018, 11:55 AM  #4 
Newbie Joined: Jan 2018 From: Iowa Posts: 1 Thanks: 0 
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.

January 23rd, 2018, 04:04 PM  #5  
Math Team Joined: Jul 2011 From: Texas Posts: 2,751 Thanks: 1401  Quote:
$v_f^2 = v_0^2  2g \cdot \Delta y$ 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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