1. Not finding help here? Sign up for a free 30min 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!

In Need of Help, Please

  1. Sep 21, 2004 #1
    Hello, I was wondering if anyone could help me out with some of the problems from my problem set. I have tried to do the ones that I will post, but I keep coming out with the incorrect answers (according to the online scoring program that we use [CAP]).

    ***On a rainy day, raindrops fall with a vertical velocity of 8.4 m/s. If a car drives through the rain at 78.0 km/hr, what is the magnitude of the velocity of the raindrops relative to the car? At what angle are the raindrops falling relative to the car? Assume the direction the car is headed is 0o and down is 90o***

    ....so for this I drew a triangle with the 'vertical' side being 8.4m/s and the 'horizontal' as -21.7 m/s (after converting from km/hr to m/s). For the first part of the question, I just did vector subtraction to find the hypoteneuse and got -13.3 m/s...but apparently that was incorrect. And then for the angle I did -tan= (21.7m/s / 8.4m/s) and got 68.81 deg...but that was wrong too. What am I doing wrong here?

    ***An athlete executing a long jump leaves the ground at a 30.0o angle and travels 7.34m. What was the take-off speed?***

    ...I thought that maybe I was looking for an x-component here and tried to solve xcos30 = 7.34 ...but that didn't work either. I think I had the wrong idea here.


    Anyway, those are just a few of the ones I'm struggling with. I know they are supposed to be easy questions...but I always tend to approach them the wrong way. Any help would be greatly appreciated. :eek:)
     
  2. jcsd
  3. Sep 21, 2004 #2
    i'll answer the second question with the athlete because i simply love projectile motion:

    Now it is projectile motion so you must split the inital velocity into two components

    Dx = 7.34m
    V1x = Vocos30
    t=?

    Dx = V1x t (because there is no external force on his horizontal)
    7.34 = (Vo cos 30 )t
    7.34/cos30 = Vo t
    Thus Vo T = 7.34/cos30
    NOW, for the Y direction, take the upward to be positive and downward to be negative

    Dy = 0 (ecasue the athlete lands on the same level)
    Ay = -9.8m/s^2 (because of gravity is in the downward direction we get negative)
    V1y = Vo Sin 30
    t = ?

    Dy = V1y t + 1/2 Ay t^2
    0 = (VoSin30) t + 0.5 (-9.8)t^2
    0 = VoT (Sin30) -4.9t^2
    Now use the identity from above in bold for Vo T

    Find T then use the missing link to find Vo

    Problem solved
     
  4. Sep 21, 2004 #3
    It worked! Thanks so much...now only a bunch of other problems left to do, lol.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: In Need of Help, Please
  1. Need help please (Replies: 4)

Loading...