The movement of a ball thrown from a ball-throwing machine forms a parabolic path given by the equation $$h(t) = -t^2 + 6t + 1,62$$ 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 find the first derivative of $h(t)$, we differentiate the function:

$h'(t) = -2t + 6$

The maximum height can be found by locating the vertex of the parabola, which occurs at

$t = -\frac{b}{2a} = -\frac{6}{2(-1)} = 3 ext{ s}$

Now, we substitute $t = 3$ into the height function:

$h(3) = -(3)^2 + 6(3) + 1,62 = 10,62 ext{ m}$

Therefore, the maximum height that the ball reaches is 10,62 m.

To find when the velocity is 3 m/s, set the derivative equal to 3:

$-2t + 6 = 3$

Solving for $t$ gives:

$-2t = 3 - 6$ $t = 1,5 ext{ s}$

Now, substituting back into the height equation:

$h(1,5) = -(1,5)^2 + 6(1,5) + 1,62 = 8,37 ext{ m}$

Thus, the height of the ball above the ground when it reaches a velocity of 3 m/s is 8,37 m.