1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Falling Cantaloupe - Distance Above Ground

  1. May 22, 2017 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations
    ## \Delta h = \frac {v_i^2 - v_f^2} {2g} ##

    3. 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?
     
  2. jcsd
  3. May 23, 2017 #2
    I think so.
     
  4. May 23, 2017 #3

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Falling Cantaloupe - Distance Above Ground
Loading...