Question

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,
(ii) (ii) the length of time for which the signal is above the level of the top of the cliff.

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)^1\2 s.
(iii) Find the speed of projection of the second

Answer 1

I got the first question how about part ii and iii

Answer 2

@e.mccormick

Answer 3

@Preetha

Answer 4

So the max height is 45m above the sea?

Answer 5

Well, if so, you can use the formula \(s=\frac{1}{2}at^2+vt\).
\(s\) is the distance between two points, in this case, it is 45m.
\(a\) is the acceleration, in this case, it is \(-g\).
And \(v\), of course, is the initial velocity.