velocity

MagicMalkin

MagicMalkin

MagicMalkin

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I'm just doing a bit of maths homework and I'm stuck! Need a push in the right direction..

A stone is thrown vertically upwards from a height of 2m above ground level. It reaches a maximum height of 5m above ground level.

I've managed to get the following.

s = 3
u = 7.67
v = 0
a = -9.8

Now I'm asked to find the velocity of the stone when it hits the ground.

How do I go about this?
 
Well you know the distance the stone falls, mavity

There is an equation which you can rearrange for v

initial velocity is zero btw
sid
 
Well if it reaches a max height of five meters it's got five meters to travel before hitting the ground.

Therefore distance traveled will be 5m, initial velocity will be zero and acceleration due to mavity will be 9.81 m/s. You can take it from there.

Edit: Beaten
 
V^2 = U^2 + 2xAS

where V is the final velocity (what you're trying to find), U is the initial velocity (zero), A is the accelleration due to mavity (9.80665m/s/s) and S is the displacement/distance (5 meters)
 
Have you not got your list of 'suvat' equations infront of you? Work out the time taken by using the speed and distance (5m) and substitute your result into the appropriate suvat formula.

edit: above is more plausible. V^2 = Initial speed^2 x (2 x acceleration of mavity x distance)
 
Last edited:
u = 0
v = ?
s = 5
a = 9.81

v^2 = u^2 + 2as

thats how you work it out

Once it reaches its max height its velocity will be 0. so u = 0
It will travel 5m before it hits the ground. so s = 5
Acceleration due to mavity = 9.81ms^2
 
Ah cheers. Had the equation but was subbing in the initial velocity of the stone being thrown upwards instead of 0. Fits in with the answer in the book now! (9.9 ms^-1)
 
hurrah! sleepy-physics is awesome! (only just woke up when posting, still blurry eyes)
 
This is far more simple than it looks. Consider it simply as falling from a height of 5 meters (since the stone will be stationary at the apex). Through conservation of energy, you just have v = sqrt(2gh) = sqrt(2*5*9.8) = sqrt(98) = about 10 m/s
 
SlyReaper said:
This is far more simple than it looks. Consider it simply as falling from a height of 5 meters (since the stone will be stationary at the apex). Through conservation of energy, you just have v = sqrt(2gh) = sqrt(2*5*9.8) = sqrt(98) = about 10 m/s
Click to expand...

which is essentially what i said, but in a confusing (read: non-standard way) which might confuse the poor chap.
 
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