The height (in meters) of a shot cannonball follows a trajectory given by $h(t) = -4.9t^2 + 14t - 0.4$ at time t (in seconds). As an improper fraction, for how long is the cannonball above a height of 6 meters?

Guest Jul 2, 2023

#1**0 **

The equation for the height of the cannonball is:

h(t) = -4.9t^2 + 14t - 0.4

We want to find the time t when the height of the cannonball is 6 meters, so we set the equation equal to 6:

-4.9t^2 + 14t - 0.4 = 6

Solving for t, we get:

t = 1/2, 4

The first solution, t = 1/2, represents the time when the cannonball is first launched. The second solution, t = 4, represents the time when the cannonball reaches a height of 6 meters for the last time.

Therefore, the cannonball is above a height of 6 meters for a total of 4 - (1/2) = 7.5 seconds.

Guest Jul 4, 2023