If a ball is thrown upward from ground level with an initial velocity
of 24 m/s, its height h in metres after t seconds is given by h (t) =
24t −16t2 . When will the ball be at a height of more than 8 metres?
To find when the ball will be at a height of more than 8 meters, we need to solve the inequality h(t) > 8.
h(t) = 24t - 16t^2
Setting h(t) > 8:
24t - 16t^2 > 8
-16t^2 + 24t - 8 > 0
-4t^2 + 6t - 2 > 0
-2(2t^2 - 3t + 1) > 0
2t^2 - 3t + 1 < 0
Solving the quadratic inequality 2t^2 - 3t + 1 < 0, we find the roots of the equation:
2t^2 - 3t + 1 = 0
(t - 1)(2t - 1) = 0
t = 1 or t = 0.5
These are the critical points. We will test the intervals created by these points in the inequality to determine when the ball is at a height of more than 8 meters.
Testing t = 0:
2(0)^2 - 3(0) + 1 = 1 > 0
Testing t = 0.5:
2(0.5)^2 - 3(0.5) + 1 = -0.75 < 0
Testing t = 1:
2(1)^2 - 3(1) + 1 = 0
Therefore, the ball is at a height of more than 8 meters for t < 0.5 and 0 < t < 1. The ball will be at a height of more than 8 meters during these time intervals.