Falling Cantaloupe - Distance Above Ground

[Catchingupquickly]

Homework Statement

A cantaloupe with a mass of 0.45 kg falls out of a tree house that is 5.4 meters above the ground. It hits a tree branch at a speed of 6.3 m/s. How high is the tree branch from the ground?

Homework Equations

$\Delta h = \frac {v_i^2 - v_f^2} {2g}$

The Attempt at a Solution

Given:

g = 9.80 $m/s^2$
inital velocity = 0
final velocity = 6.3 m/s
mass = 0.45 kg (but here it's a red herring)

$\Delta h = \frac {0 - (6.3 m/s)^2} {(2) 9.80 m/s^2} \\= \frac {-39.69} {19.6}$
= -2.025 meters

So 5.4 meters - 2.025 meters = 3.375 or 3.38 meters above the ground.

Am I correct?

 

[TomHart]

I think so.

 

[PeroK]

Homework Statement

A cantaloupe with a mass of 0.45 kg falls out of a tree house that is 5.4 meters above the ground. It hits a tree branch at a speed of 6.3 m/s. How high is the tree branch from the ground?

Homework Equations

$\Delta h = \frac {v_i^2 - v_f^2} {2g}$

The Attempt at a Solution

Given:

g = 9.80 $m/s^2$
inital velocity = 0
final velocity = 6.3 m/s
mass = 0.45 kg (but here it's a red herring)

$\Delta h = \frac {0 - (6.3 m/s)^2} {(2) 9.80 m/s^2} \\= \frac {-39.69} {19.6}$
= -2.025 meters

So 5.4 meters - 2.025 meters = 3.375 or 3.38 meters above the ground.

Am I correct?

You could always check by calculating it a different way. For example, here is a quick check:

With gravity at $9.8 m/s^2$ it takes about $0.65 s$ to reach $6.3 m/s$. If gravity were $10 m/s^2$ it would be $0.63 s$, so it's a bit more than that.

The average speed when falling from rest is half the final speed, so that's approx $3.1 m/s$.

$3.1 \times 0.65 \approx 1.95 + 0.06 = 2.01$

So, the object fell about $2 m$.