Question

A stone is dropped from the top of a cliff. Its acceleration is 9.81 ms^-2. How fast is it moving after 1 second, and after 3 seconds?

Answer 1

s=ut+1/2 at^2

Answer 2

u=0, so its just s=1/2at^2

Answer 3

a=9.8 and t=1 or 3

Answer 4

I set this one up for the first part as 9.81 = the change in velocity over 1 second so I got 9.81 for part 1.

Answer 5

yes

Answer 6

But with your formula wouldn't it be 4.9?

Answer 7

Yup! :)

Answer 8

\[a=\frac{ dv}{dt}\]

Answer 9

Then for the second part I got 9.81 = x over 3. So i Multiplied 3 * 9.81 and got 29.43 for how fast its going after 3 seconds.

Answer 10

@caozeyuan , your formula was for distance!

Answer 11

Yup, @tombraider5000 ! You've got it.

Answer 12

\[v=\int adt\]

Answer 13

since a is const, v=at

Answer 14

Thanks for the help caozeyuan and eric :)

Answer 15

You're welcome :)

Answer 16

Good job!