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Homework Help: Physics Problem:Kinetic Energy

  1. Oct 12, 2004 #1
    Can someone please tell me what I did wrong for Part A. I used this equation: KE=1/2m*v^2. The answer I got was 58.8J but its wrong. Am I using the right equation?

    A 47.0 g golf ball is driven from the tee with an initial speed of 50.0 m/s and rises to a height of 23.4 m.
    (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. ?J

    (b) What is its speed when it is 7.0 m below its highest point?
  2. jcsd
  3. Oct 12, 2004 #2
    how fast is the ball going when it is at Hmax?
    this is the v you need to use.
  4. Oct 12, 2004 #3
    how fast is the ball going when it's at Hmax is actually the horizontal velocity
    vx = vo x cosa (where vo is the initial velocity and a is the angle of projection)
    I think the additional needed info in this question is the angle of projection.

  5. Oct 12, 2004 #4
    I hope this site may help you Shawonna
    http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Vectors/ProjectilesMotion.html [Broken]

    Last edited by a moderator: May 1, 2017
  6. Oct 12, 2004 #5

    They are making this a bit confusing for you i think. Keep in mind this is a two dimensional problem, so you need to calculate the x and y components of the velocity v by using the well known triangle-rules with sine and cosine.

    At the highest point, the y-component of the velovity is 0 because you do not move upwards anymore. So v_y = 0

    The x-component is constant v_x = v' * cos (a) where a is angle of inclination of the ball with respect to the x-axis. The v' is the given initial velocity. What you need to find is the angle a. Try using this t = v'_y/g is the time that the y component of the velocity is 0. This t-value must be substituted into the expression for the y-component of the position (max height) : y - y' = v'_y * t - gtĀ²/2. And y-y' = 23.4. Now you can calculate v'_y and you know v'. Via v'_y = v' * sin (a) you will find a.

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