Question

3 The top of a cliff is 40 metres above the level of the sea. A man in a boat, close to the bottom of the cliff, is in difficulty and fires a distress signal vertically upwards from sea level. Find
(i) the speed of projection of the signal given that it reaches a height of 5 m above the top of the cliff, [2]
(ii) the length of time for which the signal is above the level of the top of the cliff. [2] The man fires another distress signal vertically upwards from sea level. This signal is above the level
of the top of the cliff for ﰆﰄ17ﰅ s.
(iii) Find the speed of projection of the second sig

Answer 1

you can use the equations of motion with constant acceleration here.
for example
(I) displacement s = 50 + 5 = 45 m, final velocity v = 0, acceleration due to gravity g = -9.81 ms-2, initial velocity u = ?

use v^2 = u^2 + 2gs

0 = u^2 -2 * 9.81 * 45

solve for u