a stone is dropped down a well and strikes the bottom of the well 4s later. the acceleration due to gravity is 10m/s

Question

anonymous · Answer

How deep is the wall?

anonymous · Answer

4 * 10m/s = _?__ m

anonymous · Answer

solve this problem using vertical components 
plug in all your information to one of the distance formulas $$\Delta dy = Viy \Delta t + \frac{ 1 }{ 2 } a ( \Delta t) ^2$$ but use your Vi or initial velocity as 0 since the ball was "dropped" not thrown

anonymous · Answer

@ihatephysicslol good name and yea thats correct. the m/s threw me off and not m/s2

anonymous · Answer

lol thankyou, and it happens haha :)

anonymous · Answer

stupid mistake :(

shamim · Answer

$$h=ut+\frac{ 1 }{ 2 }g t ^{2}=\frac{ 1 }{ 2 } g t ^{2}=\frac{ 1 }{ 2 } \times 10 \times 4^{2}$$