The function h=-5t^2 + 20t + 2 gives the approximate height, h metres, of a thrown football as a function of time, t seconds, since it was thrown. The ball hit the ground before the receiver could catch it. For how many seconds was the height of the ball at least 17 m?
-
we want to let h = 17
-5t^2 + 20t + 2 = 17
-5t^2 + 20t - 15 = 0 --- divide by -5 to simplify the equation
t^2 - 4t + 3 = 0
(t - 1)(t - 3) = 0
t = 1, t = 3
what this means is it was at 17m on the way up at 1s, and at 17m on the way down at 3s, so the total time spent above 17m is:
t = 3 - 1
t = 2s
You can use this method or sketch the curve by a graphics calculator or otherwise and find the time interval when the graph is above 17m.
-5t^2 + 20t + 2 = 17
-5t^2 + 20t - 15 = 0 --- divide by -5 to simplify the equation
t^2 - 4t + 3 = 0
(t - 1)(t - 3) = 0
t = 1, t = 3
what this means is it was at 17m on the way up at 1s, and at 17m on the way down at 3s, so the total time spent above 17m is:
t = 3 - 1
t = 2s
You can use this method or sketch the curve by a graphics calculator or otherwise and find the time interval when the graph is above 17m.