Homework Help: 1 Q need help

1. Mar 31, 2014

nouf33

1. The problem statement, all variables and given/known data
a stone is thrown vertically downward from the top of 40 m tall building with an initial speed of 1 after 2s the stone will have traveled a distance of?

2. Relevant equations
x=vt+0.5at^2

vf=vi+at

3. The attempt at a solution
1(2)+0.5(-9.8)(2)^2=-17.6

the answer i was given is 22

and i tried to put 9.8 without negative and it was 21.6 is this one right ??
why is the acceleration positive while going up is negative ???

waiting and thank you

2. Mar 31, 2014

BvU

Hello Nouf, and welcome to PF.
The key is in your inital speed. (Don't forget to always add units to values if it is for a variable with a dimension, like here: v = 1 m/s)

Speed (velocity) has a magnitude and a direction. It matters if the stone is thrown up [ meaning v(0) = 1 m/s ] or down [ meaning v(0) = -1 m/s ]. This orientation I copied from your using (-9.8) as the acceleration, which is definitely towards the earth, so downwards!

The reason the answer is 21.6 and not -21.6 is because usually distance traveled is taken to be positive. In this case from y = 0 (top of building) to y = -21.6 Distance is |y(2) - y(0)|.

It is a good habit ("natural choice") to choose coordinates such that x is horizontal and y is vertical in 2 dimensional problems. For 3 dimensional problems we usually pick x and y in the horizontal plane and z up. (x to the right, y forward).