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

To find the initial velocity of the toy rocket, we first compute the derivative of the height function:

$h'(t) = \frac{dh}{dt} = -10t + 25$

Now, substituting t = 0 for initial velocity:

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

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

To find the maximum height, we first determine when the rocket reaches its peak by setting the derivative equal to zero:

$-10t + 25 = 0$

Solving for t:

$t = \frac{25}{10} = 2.5 \, \text{s}$

Now we will substitute t = 2.5 back into the height function:

$h(2.5) = -5(2.5)^2 + 25(2.5) = -5(6.25) + 62.5 = -31.25 + 62.5 = 31.25 \, \text{m}$

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

To determine when the rocket reaches 30 m, we set the height function equal to 30:

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

Rearranging gives us:

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

Dividing through by -5 yields:

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

Factoring the quadratic:

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

This gives us the solutions:

$t = 2 \, \text{s} \quad \text{and} \quad t = 3 \, \text{s}$

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