Physics

The projectile launcher shown below will give the object on the right an inital horizontal speed of 9.40 m/s. While the other object will be dropped with no initial speed. The objects are initially 114 cm above the ground. What will be the difference in the speeds of the two objects when they reach the ground?

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  1. I will have to guess at your drawing.
    Both objects fall the same.
    The first one keeps its horizontal velocity
    so
    both fall 1.14 meters
    v = g t = 9.81 t

    1.14 = (1/2) g t^2 = 4.9 t^2
    so t = 0.482 second fall time for both
    so vertical velocity of both is
    9.81 * .482 = 4.73 m/s

    that is the entire velocity of the dropped one

    for the projectile
    speed = sqrt(4.73^2 + 9.4^2)

    so calculate that and subtract 4.73 to get the speed difference

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