   SEARCH HOME Math Central Quandaries & Queries  Question from Reuben, a student: A ball is h meters above the ground t seconds after it is thrown where h(t) = 40t - 5t ^2 . Find the time at which the ball first reaches a height of (a) 10 meters (b) 40 meters (c) 100 meters Reuben,

For (a) you are to find the time when h(t) = 10 meters. Substituting into the height expression gives

10 = 40t - 5t2

On simplification this becomes

t2 - 8t + 2 = 0.

Solve for t (you will probably have to use the general quadratic formula).

• If you get 2 real solutions then the ball is at 10 meters twice, once on the way up and again on the way down.

• If you get no real solutions then the ball never reaches that height.

• If you get exactly one solution then this is the time when the ball reaches its maximum height and begins it trip back down.

Apply the same analysis to parts (b) and (c).

Harley     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.