| Date | May 2011 | Marks available | 3 | Reference code | 11M.2.SL.TZ2.2 |
| Level | Standard level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Determine | Question number | 2 | Adapted from | N/A |
Question
This question is about kinematics.
Lucy stands on the edge of a vertical cliff and throws a stone vertically upwards.
The stone leaves her hand with a speed of 15ms–1 at the instant her hand is 80m above the surface of the sea. Air resistance is negligible and the acceleration of free fall is 10ms–2.
Calculate the maximum height reached by the stone as measured from the point where it is thrown.
Determine the time for the stone to reach the surface of the sea after leaving Lucy’s hand.
Markscheme
\(h = \frac{{{v^2}}}{{2g}}\);
\( = \left( {\frac{{225}}{{20}} = } \right)11{\rm{m}}\);
Award [1 max] for 91m or 91.25m (candidate adds cliff height incorrectly).
time to reach maximum height=1.5s;
time to fall 91m=4.3s;
total time=5.8s;
Answer can be alternatively expressed as 3.0 (to return to hand) +2.8 (to fall 80m).
or
use of s=ut+½at2;
80=-15t+5t2 or −80=15t−5t2;
t=5.8s;
