SOLUTION: a toy rocket is launched from the top of a building 50 ft. tall at an initial velocity of 90 ft per second. Let t represent the amount of time elapsed after the launch. Solve.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: a toy rocket is launched from the top of a building 50 ft. tall at an initial velocity of 90 ft per second. Let t represent the amount of time elapsed after the launch. Solve.       Log On


   

Question 104934: a toy rocket is launched from the top of a building 50 ft. tall at an initial velocity of 90 ft per second. Let t represent the amount of time elapsed after the launch. Solve.

the answer is y= -16x2 + 90x + 50
How did the get the -16x squared??
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Well...
The general form of the equation that gives the height (h) of an object propelled upwards, as a function of time (t) is:
h%28t%29+=+-%281%2F2%29gt%5E2%2Bv%5B0%5Dt%2Bh%5B0%5D where: h is the height in feet, t is the time in seconds, g is the acceleration due to gravity (g+=+32+ft%2Fsec%5E2%29, v%5B0%5D is the initial velocity of the object, and h%5B0%5D is the initial height of the object.
If you wanted to graph such an equation (it would be a parabola opening downward), it would be written as:
y+=+-16t%5E2%2Bv%5B0%5Dt%2Bh%5B0%5D
The x in your equation is a replacement variable for the time, t.
The first term -16t%5E2 is negative because of the downward effect of gravity on the object.
Just for fun, let's see what your equation would look like when graphed:
graph%28600%2C400%2C-5%2C8%2C-5%2C200%2C-16x%5E2%2B90x%2B50%29