Question

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

Answer 1

How deep is the wall?

Answer 2

4 * 10m/s = _?__ m

Answer 3

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

Answer 4

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

Answer 5

lol thankyou, and it happens haha :)

Answer 6

stupid mistake :(

Answer 7

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