Question

A man stands on a cliff that 10m above the sea. He throws a stone vertically upwards with velocity 5ms^-1. The stone eventually lands in the sea, if the air resistance is negligible, what is the time taken for the stone to reach the sea after leaving the man hand.

Answer 1

Do you remember the general equation for the distance covered
by a moving object ?  There are not many perfect opportunities to
use it in all its glory, but I think this is one of them.

Position =
(starting distance) + (starting speed) (time) + (1/2) (acceleration) (time)²

H = starting position + (starting speed x t) + 1/2 A t²

Here's how we can use it, with some careful definitions:

-- Let's say the surface of the sea is zero height.
Then 'H' ... the position at the end ... is zero, when it plunks into the water, and
the starting, original position of the stone is +10 on the cliff in the man's hand.

--  Starting speed is +5 ... 5 m/s upward, when he tosses it.

-- Acceleration is 9.8 m/s² downward ... the acceleration of gravity.

I think this is going to work out just beautifully !

0 = (5) + 5t - 1/2 (9.8) t²

-4.9 t² + 5t + 5 = 0  That's the whole thing right there. Look how gorgeous that is !

Solve it for 't' with the quadratic equation,
A = -4.9
B = 5
C = 5

When you solve a quadratic with the formula, you always get two roots.
If it's a real-world situation, one of them might not make sense.  That's
the result in this case.

The two roots are

t = - 0.622 second
and
t = + 1.642 second

The first one isn't useful, because it means 0.622 second before the man
tossed the stone up.

So our answer is:  We hear the 'plunk' 1.642 second after the upward toss.