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# Time and velocity Watch

1. A stone falls down a well vertically. Ignore friction.

it falls 60m , initial speed = 5m/s

acceleration must be -9.8m/s/s

If i use s = ut +1/2at^2 I can get an answer but it comes out as

-60 = 5t - 4.9t^2

I'm fairly sure you shouldn't have to use quadratic formula for a short question. What's wrong?
2. What's the issue with using the quadratic formula?

You've messed up a sign btw.
3. (Original post by alow)
What's the issue with using the quadratic formula?

You've messed up a sign btw.
got the mistake.My point was is there a different way
4. (Original post by Feynboy)
which one?
If you're saying the fall is in the -ve direction, what is your starting velocity?
5. The quadratic form would be
0 = - 4.9t^2 + 5t + 60
6. (Original post by Feynboy)
A stone falls down a well vertically. Ignore friction.

it falls 60m , initial speed = 5m/s/s

acceleration must be -9.8m/s/s

If i use s = ut +1/2at^2 I can get an answer but it comes out as

-60 = 5t - 4.9t^2

I'm fairly sure you shouldn't have to use quadratic formula for a short question. What's wrong?
Acceleration is positive if it's falling down. The equation should be:-
4.9t^2 + 5t-60=0.
7. (Original post by sabahshahed294)
Acceleration is positive if it's falling down. The equation should be:-
4.9t^2 + 5t-60=0.
Not if you take upwards as +ve. As long as you're consistent it can be either way.
8. (Original post by alow)
Not if you take upwards as +ve. As long as you're consistent it can be either way.
You mean the displacement right? Because if a=g here then a is positive.
9. (Original post by sabahshahed294)
You mean the displacement right? Because if a=g here then a is positive.
I mean taking upwards as being +ve displacement, like OP has done.

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