2 A box of emergency supplies is dropped to victims of a natural disaster from a stationary helicopter at a height of 1000 metres. The initial velocity of the box is zero. At time \(t \mathrm {~s}\) after being dropped, the acceleration, \(a \mathrm {~m} \mathrm {~s} ^ { - 2 }\), of the box in the vertically downwards direction is modelled by $$\begin{aligned} & a = 10 - t \text { for } 0 \leqslant t \leqslant 10
& a = 0 \quad \text { for } \quad t > 10 \end{aligned}$$

1. Find an expression for the velocity, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), of the box in the vertically downwards direction in terms of \(t\) for \(0 \leqslant t \leqslant 10\). Show that for \(t > 10 , v = 50\).
2. Draw a sketch graph of \(v\) against \(t\) for \(0 \leqslant t \leqslant 20\).
3. Show that the height, \(h \mathrm {~m}\), of the box above the ground at time \(t\) s is given, for \(0 \leqslant t \leqslant 10\), by $$h = 1000 - 5 t ^ { 2 } + \frac { 1 } { 6 } t ^ { 3 }$$ Find the height of the box when \(t = 10\).
4. Find the value of \(t\) when the box hits the ground.
5. Some of the supplies in the box are damaged when the box hits the ground. So measures are considered to reduce the speed with which the box hits the ground the next time one is dropped. Two different proposals are made. Carry out suitable calculations and then comment on each of them.
   (A) The box should be dropped from a height of 500 m instead of 1000 m .
   (B) The box should be fitted with a parachute so that its acceleration is given by $$\begin{gathered} \quad a = 10 - 2 t \text { for } 0 \leqslant t \leqslant 5 ,
   a = 0 \quad \text { for } \quad t > 5 . \end{gathered}$$