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energy

Each minute 12 000 kg of waterfalls vertically into the pool at the bottom. The time taken for the water to fall is 2 s and the acceleration of the water is 10 m/s^2. (a) Assume the speed of the water at the bottom of the dam is zero. Calculate the speed of the water just before it hits the pool at the bottom.
(edited 5 years ago)
Seems to be asking what the speed of an object is after accelerating at constant rate of 10ms-2 for 2 seconds

you could solve that with SUVAT
Reply 2
Avatar for Fa.ara
Fa.ara
OP
12
Original post by Joinedup
Seems to be asking what the speed of an object is after accelerating at constant rate of 10ms-2 for 2 seconds

you could solve that with SUVAT


what's suvat
Reply 3
Avatar for Kane987
Kane987
7
SUVAT is used for calculations when the acceleration is constant.
It stands for:

S - Displacement
U - Initial Velocity
V - Final Velocity
A - Acceleration
T - Time taken

There are 4 equations you can use from them which are:

v = u + at
s = (v+u)/2 x t
s = ut + 0.5at^2
v^2 = u^2 + 2as

In your question specifically:

'The time taken to fall is 2s' (so t = 2)
'The acceleration of the water is 10ms^-2' (so a = 10)

However, I don't know if I'm being silly, but I think you can assume that u = 0 because there was no "vertical velocity" before the water started falling (?),
so 'v = u + at' becomes 'v = at',

so you do v = 10 x 2 = 20ms^-1
(edited 5 years ago)

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