Motion Under GravityDate: 07/18/2003 at 22:30:38 From: Jen Subject: Algebra Motion Under Gravity: Use the equation s = -16t squared + vt to determine when the height of an object is 960 feet if the initial velocity is 272 feet per second. Solve either algebraically or graphically. Include all algebra to justify response if solving algebraically. Include sketch of graph and explain how you interpreted the graph to get your answer(s). Date: 07/21/2003 at 12:56:42 From: Doctor Ian Subject: Re: Algebra Hi Jen, The equation is s = -16t^2 + vt Let's think about what this means. It means that at time zero, the height is also zero. So the object is starting out from the ground, or wherever we're measuring height from. Now, suppose there were no gravity. Then we could figure out how high the object is by multiplying the velocity by the time. This is the same thing we do when we note that at 60 miles per hour, a car travels 60 miles in one hour, 120 miles in two hours, and so on. But there _is_ gravity, and that's going to accelerate the object back toward earth. That's why it has a negative sign, because it's working in a direction opposite to the initial velocity. Note that as t increases, the size of the acceleration term will eventually become larger than the size of the velocity term (because t^2 will eventually be much larger than t), so eventually the object will end up returning to the ground. So that's what the equation is saying, albeit in a very compact way. You're told that the initial velocity is 272 feet/sec. So you can substitute that for v: s = -16t^2 + 272t And you're told that at some point, the height will be 960 feet: 960 = -16t^2 + 272t which means that 16t^2 - 272t + 960 = 0 And dividing both sides of the equation by 16 gives us t^2 - 17t + 60 = 0 So this is a quadratic equation with variable t. You can try to factor it, or you can use the quadratic formula to get the answer directly. Is this enough to get started? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/