SOLUTION: A ball is tossed upward with an initial velocity of 122 ft/s from a platform that is 700 ft above the surface of the earth. After t seconds, the height of the ball above the ground

Click here to see ALL problems on Geometry Word Problems

Question 732251: A ball is tossed upward with an initial velocity of 122 ft/s from a platform that is 700 ft above the surface of the earth. After t seconds, the height of the ball above the ground is given by the equation h = -16t^2 + 122t + 700. What is the maximum height of the ball? Round to the nearest tenth of a foot.
Found 2 solutions by lynnlo, ikleyn:
Answer by lynnlo(4176) (Show Source):
You can put this solution on YOUR website!

Answer by ikleyn(53956) (Show Source):
You can put this solution on YOUR website!
.
A ball is tossed upward with an initial velocity of 122 ft/s from a platform that is 700 ft above
the surface of the earth. After t seconds, the height of the ball above the ground is given
by the equation h = -16t^2 + 122t + 700. What is the maximum height of the ball?
Round to the nearest tenth of a foot.
~~~~~~~~~~~~~~~~~~~~~~

This given equation h = -16t^2 + 122t + 700 has the leading coefficient negative, -16. So, it describes a parabola opened downward. Such a parabola has a maximum. According to the general theory, a parabola y = ax^2 + bx + c with a negative leading coefficient 'a' has a maximum at the point x = . In our case, the maximum is achieved at x = = = 3.8125. It means that the maximum height is achieved at 3.8125 second after tossing. To find the maximum height h, substitute x = 3.8125 into the formula and calculate h = -16*3.8125^2 + 122*3.8125 + 700 = 932.6 feet (rounded as requested). ANSWER. The maximum height of the ball is 932.6 feet.

Solved.