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October 22nd, 2009, 06:19 PM  #1 
Member Joined: Sep 2009 Posts: 51 Thanks: 0  Differential Equation: Motion Problem
A canyon is 1100 meters deep at its deepest point. A rock is dropped from the rim above this point. Write the height of the rock as a function of time t in seconds. (Use 9.8 m/s2 as the acceleration due to gravity.) h(t) = How long will it take the rock to hit the canyon floor? (Give your answer correct to 1 decimal place.) ________s ok so for the first part I got h(t)= 9.8t+1100 do i take the integral of that? or what steps do i take? Thanks 
October 22nd, 2009, 10:55 PM  #2 
Senior Member Joined: Jan 2009 From: Japan Posts: 192 Thanks: 0  Re: Differential Equation: Motion Problem
Start with acceleration: a(t) = 9.8 Integrate this to get velocity: v(t) = 9.8t + v0 where v0 is your initial velocity. Here the word "dropped" implies v0 = 0, so v(t) = 9.8t. Integrate this to get position: h(t) = 4.9t^2 + h0 where h0 is your initial height. Here, you are told it is 1100 meters, so h(t) = 4.9t^2 + 1100 To find the time it takes the hit the canyon floor, let h(t) = 0 and find t. 
October 24th, 2009, 11:38 AM  #3 
Member Joined: Sep 2009 Posts: 51 Thanks: 0  Re: Differential Equation: Motion Problem
thank you so much for explaining...this really helped


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