A toy rocket is launched upwards from the ground - NSC Technical Mathematics - Question 8 - 2022 - Paper 1

The initial velocity can be found by calculating the derivative of the height function, which gives the velocity function:

$h'(t) = -10t + 25$

For the initial velocity, evaluate at t = 0:

$h'(0) = -10(0) + 25 = 25 \, m/s$

Therefore, the initial velocity of the toy rocket is 25 m/s.

To find the maximum height, first determine the time when the height is at its peak. This occurs when the velocity is zero:

Set the velocity function to zero:

$-10t + 25 = 0$

Solving for t gives:

$t = 2.5 \, s$

Now substitute this value back into the height function:

$h(2.5) = -5(2.5)^2 + 25(2.5)$

Calculating gives:

$h(2.5) = -5(6.25) + 62.5 = -31.25 + 62.5 = 31.25 \, m$

Thus, the maximum height that the toy rocket reached is 31.25 m.

Set the height equation equal to 30 m:

$30 = -5t^2 + 25t$

Rearranging gives:

$0 = -5t^2 + 25t - 30$

Dividing by -5 results in:

$0 = t^2 - 5t + 6$

Factoring the quadratic:

$0 = (t - 2)(t - 3)$

Thus, the solutions are:

$t = 2 \, s \, ext{and} \, t = 3 \, s$

Therefore, the values of t for which the rocket will be 30 m above the ground are 2 s and 3 s.