The movement of a ball thrown from a ball-throwing machine forms a parabolic path given by the equation h(t) = -t² + 6t + 1,62 where h is the height (in metres) of the ball above the ground and t is the time in seconds - NSC Technical Mathematics - Question 8 - 2024 - Paper 1

To determine the derivative of the height function with respect to time, we differentiate the height equation:

h'(t) = -2t + 6$$ This expression gives us the instantaneous rate of change of height at any time t.

To find the maximum height, we first set the derivative equal to zero to find the critical points:

h'(t) = 0 \ -2t + 6 = 0 \ t = 3$$ Now we substitute t = 3 back into the height equation:

h(3) = -(3)^2 + 6(3) + 1.62 = 10.62 ext{ m}$$

Therefore, the maximum height the ball reaches is 10.62 meters.

To find when the ball reaches a velocity of 3 m/s, we set the derivative equal to 3:

h'(t) = 3 \ -2t + 6 = 3 \ -2t = -3 \ t = 1.5$$ Next, we substitute t = 1.5 into the height equation to find the height:

h(1.5) = -(1.5)^2 + 6(1.5) + 1.62 = 8.37 ext{ m}$$

Thus, the height of the ball above the ground when it reaches a velocity of 3 m/s is 8.37 meters.