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 (17 )s.
(iii) Find the speed of projection of the second sign

Answer 1

\[s = ut + 1/2 a t^2\]
s = 45m
u = initial speed = ??? m/s
v = final speed = 0 m/s
a = gravity = -9.8 m/s^2
t = ?

... hmm, u dont have t, so that wont work

try:
\[v^2 = u^2 + 2as\]
some constants and variables as above

Answer 2

*same... not some

Answer 3

@LojynLoay ...?

Answer 4

This is the question link
http://olevel.sourceforge.net/papers/9709/9709_s13_qp_41.pdf
it is question 3 . I don't know how to solve it and the answer link is
http://olevel.sourceforge.net/papers/9709/9709_s13_ms_41.pdf

Answer 5

k, have u seen the kinematics equation before...?

Answer 6

because if you havent seen these equations before you probably havent done physics, and if you havent done physics im not sure how you'd be expected to know these...?