Stone throw find initial speed

[nouf33]

Homework Statement

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?

Homework Equations

x=vt+0.5at^2

vf=vi+at

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

 

[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).