Free-Fall Physics Homework: Find Velocity After 2s

[5minutes]

Homework Statement

If a boy dropped a rock off a high cliff, how fast would it be falling after 2.0 seconds had elapsed? (ignore air resistance)

a) -2.0 * 10^1 m/s
b) 2.0 * 10^2 m/s
c) 35 m/s
d) -35 m/s

Homework Equations

vf = vi + g∆t

vf = final velocity

vi = initial velocity

g = gravity = -9.81 m/s^2

∆t = change in time

The Attempt at a Solution

since the initial velocity is zero,

the velocity after 2 seconds is

V(t) = 0 + gt,

V(t) = gt,

V(2) = (9.8 m/s2)(2s),

V(2) = 19.6 m/s

The answer in my book says it is a) -2.0 * 10^1 m/s
How do you get that answer?

 

[rl.bhat]

If you take g negative, V(2) = -gt.
Apply the significant number rule to get the result.

 

[5minutes]

If you take g negative, V(2) = -gt.
Apply the significant number rule to get the result.

I'm not too sure on how to apply the signif. number rule; could you please elaborate

 

[rl.bhat]

I'm not too sure on how to apply the signif. number rule; could you please elaborate

19.6 m/s can be written a 20.0 m/s.

 

[rwisz]

The answer in your book is correct. It is written in scientific notation, which is a way of expressing very large or very small numbers in a more compact form. In this case, -2.0 * 10^1 m/s can be rewritten as -20 m/s.

To understand how this answer was obtained, we can use the formula vf = vi + g∆t. As you correctly stated, the initial velocity (vi) is 0 m/s. The value for gravity (g) is -9.81 m/s^2, which is negative because it represents the acceleration due to gravity pulling the rock downwards. The change in time (∆t) is 2 seconds.

Plugging these values into the formula, we get:

vf = 0 + (-9.81 m/s^2)(2 s)
vf = -19.62 m/s

Since the final velocity is negative, this means that the rock is falling downwards. The magnitude of the velocity is 19.62 m/s, which is approximately 20 m/s. This is where the -2.0 * 10^1 m/s answer comes from - it is just a different way of writing the same value.