More Calc
posted by Tezuka .
There is a ditch that is 18 feet deep at its deepest point. A rock is dropped from the rim above the deepest point.
a. Write the height of the rock as a function of t in seconds.
b. How long will it take for the rock to hit the ditch floor.
We want a position function, or y(t). If we use the standard equation s=(1/2)g*t^2 for the distance fallen in time tand substitute y for s we have
y(t)=18(1/2)g*t^2
To see how long it takes to hit the floor, solve for y(t)=0, or 0=18(1/2)g*t^2
g=32ft/sec^2
Why does g=32ft/sec^2?
Is that just a standard equation?
Because you stated the original problem in feet. Also, g is the acceleration due to gravity, and yes, that is a standard equation from physics.
We typically state s=(1/2)a*t^2 where s is the distance moved, a is the acceleration and t is the time. It's probably better to keep the fraction 1/2 separate from the acceleration constant to, you'll see why when you start integrating and differentiating these functions. The first derivative of this equation is s'(t)=a*t which is the equation for the veocity at time t. The second derivative s"(t)=a gives the constant of acceleration. If you're given the constant of acclereation and asked to find the velocity you would integrate a to get s'(t)=at + C for the indefinite integral.
If your question had been given in meters, then we'd use g=9.8m/sec^2
So if the function were in meters, the full equation would be...
0 = 18(1/2)(9.8)t^2, right?
And how does the live help on this site work?
That would be correct. As for the live help, I'm not sure if it's working right now.
check that, you would need to convert the 18ft to meters, then the equation would be correct.
Well, that was assuming that the problem itself was in meters, but I see your point.
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