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Ideal Projectile Motion , horizontal and vertical components?

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data

    You throw a ball toward a wall with a speed of 32 m/s at an angle of 40.0° above the horizontal directly toward a wall. The wall is 22.0 m from the release point of the ball.


    (a) How far above the release point does the ball hit the wall?


    (b) What are the horizontal and vertical components of its velocity as it hits the wall?


    3. The attempt at a solution

    a).
    (Ux)* t = Sx
    (32cos40)* t = 22
    t = 22/(32cos40)

    Sy = (Uy)t+0.5(Ay)t^2
    = {(32sin40)*[22/(32cos40)]
    +0.5*(-9.8)*[22/(32cos40)]}m
    = 12.16m i know this is wrong


    b)b).
    Vx= 32cos40
    = 24.513m/s this is correct


    (Vy)^2 = (Uy)^2+2(Ay)*(Sy)
    = (32sin40)^2+2*(-9.8)*(-12.16)
    Vy = 25.72m/s this is wrong

    having problems with vertical component and part A to the question
     
  2. jcsd
  3. Oct 26, 2009 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    You correctly found the time it hits the wall: t = d/vx = 22/32 cos40

    You used the correct formula for height: h = h0 + vyt - .5gt^2 but you forgot to square t!

    AM
     
  4. Oct 26, 2009 #3
    a).
    (Ux)* t = Sx
    (32cos40)* t = 22
    t = 22/(32cos40)

    Sy = (Uy)t+0.5(Ay)t^2
    = {(32sin40)*[22/(32cos40)]
    +0.5*(-9.8)*[22/(32cos40)]^2}m
    -4.9*1.0626
    = -5.20679
    this still seems to be wrong
     
  5. Oct 26, 2009 #4

    Andrew Mason

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    Science Advisor
    Homework Helper

    I get 14.5 m:

    t = 22/32cos40 = 22/24.5 = .90 sec.

    h = vyt - .5gt^2 where vy = 32sin40= 20.5 m/s
    = 20.5(.90) - .5 x 9.8(.9^2)
    = 18.5 - 4.0
    = 14.5 m

    AM
     
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