SOLUTION: An object is throwen from the origin of a coordinate system with the x-axis along the ground and the y-axis vertical. Its path, or trajectory, is given by the equation y=400x-16 x

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: An object is throwen from the origin of a coordinate system with the x-axis along the ground and the y-axis vertical. Its path, or trajectory, is given by the equation y=400x-16 x       Log On


   

Question 10673: An object is throwen from the origin of a coordinate system with the x-axis along the ground and the y-axis vertical. Its path, or trajectory, is given by the equation y=400x-16 x squared. Find the objects maximum height.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
y+=+400x+-+16%5E2
This is the equation of the height of an object propelled upward at an intial velocity of 400 ft/sec at an initial height of zero ft. It is usually written as:
h+=+-16t%5E2+%2B+400t Where: h = height and t = time.
The maximum height occurs at the vertex of the parabola whose x-coordinate (or t-value) is given by: -b%2F2a. This value of t will be the time at which the object reaches its maximum height. The equation must be in the standard form of:
ax%5E2+%2B+bx+%2B+c or, in this case: at%5E2+%2B+bt+%2B+c But here, a = -16, b = 400, and c = 0.
So, %28-b%2F2a%29+=+%28-400%2F%282%28-16%29%29%29 = 12.5 secs. The time at maximum height.
Now, we substitute this value of t into the original eqation and solve for h to get the maximum height.
-16%2812.5%29%5E2+%2B+400%2812.5%29+=+-2500+%2B+5000 = 2500 feet. This is the maximum height attained by the object.