a ball is given a velocity of 3.89 m/s at an angle of 32 degrees. Calculate the maximum height and the range of the ball after 5.0s.

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asked by Sam
  1. Kinematics time:

    First you need to realize that x and y act independently of each other. If you drop a ball straight downwards, it's not going to suddenly accelerate in the x direction.

    Next draw a free body diagram. The only force that is theoretically acting on it is gravity.

    Acceleration caused by gravity is equal to 9.8 m/s^2.

    Now write kinematic equations:


    Xo (initial position) is 0, so ignore it in this case. This is because the ball is launched and lands on the same elevation.

    Vo is 3.89m/s, but you need to divide it into its x and y components, so you'll actually get a Vox and a Voy.

    Acceleration is -9.8m/s^2 since gravity acts in a downward motion. This is only in the y direction, so the ball won't experience any acceleration in the x direction.

    Time links both equations together since the ball can't be in two places at the same time.

    It's relatively easy to calculate the range (I assume the change in x distance) right now using the kinematic equation. Since there's no acceleration in the x direction, we assume it's 0, therefore you get the equation

    Xx(t) = Vox*t

    Vox is initial velocity in the x direction, which we can find by using the cos function. 3.89*cos(32) will return Vox. Multiple Vox by 5s and you get the range of the ball after 5 seconds.

    As for maximum height, we use a different equation.

    Velocity final = velocity initial + acceleration* time.

    At the maximum height, the ball has a velocity of 0m/s, therefore we set Vfinal to 0.
    Voy can be found using the same method for Vox, except use the sine function.

    Voy = 3.89*sin(32).

    Gravity is -9.8 m/s^2 in the y direction. Now we plug and chug to solve for the time after the ball is launched when it reaches the maximum height.

    From there we plug in the time that we just solved for into the y kinematic equation.

    Xy = Xoy + Voy*t + 1/2 at^2

    All variables are known and we solve for Xy which is the height at a given time (in this case the time of max height).

    Sorry my writing is sort of scattered

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