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Problem check please (projectile motion)

  1. Oct 1, 2003 #1
    Hi all,

    I just wanted to check up on a problem I just did.


    The problem is:
    A vall is thrown by a baseball player. The ball is released from the player's hand at an angle of 35 degrees above the horizontal direction at a height of 2.0 meters above the playing field. The ball lands on the field 3 seconds later after it leaves the players hand. What is the inital speed of the ball as it leaves the players hand

    This is how I attempted to solve the problem, but not sure if it is correct or not.

    known variables
    y=2.0m
    t=3.00s
    a=-9.8m/s^2
    angle = 35 degrees

    first, find the inital speed in the y direction:
    y=vo(t)+.5(g)(t)^2
    2.0m=vo(3.00s)+.5(-9.8)(3.00s)^2
    v=15.37 m/s

    now make this a y-component vector and find the x component of this vector by taking the Vi of y and dividing it by (tan 35). I get 20.39 m/s. Now since the speed is the magnitude of the velocity, take the magnitude of both the x and y components of the Vi vector and I get:
    25.35m/s.

    Any help would be greatly appreciated.
     
  2. jcsd
  3. Oct 2, 2003 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    "first, find the inital speed in the y direction:
    y=vo(t)+.5(g)(t)^2
    2.0m=vo(3.00s)+.5(-9.8)(3.00s)^2
    v=15.37 m/s"

    This is the initial vertical speed necessary for the ball to reach a height of 2.0 m ABOVE the ballplayers hand. If the ball was thrown from a height of 2.0 m above the ground, then it will hit the ground when v0(3 s)+ (1/2)(-9.8)3^2= -2.
    (That's the same as taking the ground to be h=0, the initial height of the ball 2 m so you are solving 2+ v0(3)+ (1/2)(-9.8)3^2= 0.)
     
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