Physics - 2-D kinematics

posted by Anonymous

A soccer ball is kicked with an initial horizontal velocity of 20 m/s and an initial vertical velocity of 16 m/s.

I've calculated:

Initial Speed: 25.61 m/s
Angle: 38.66 degrees
Maximum Height it reaches: 13.06 m
Distance kicked: 65.29m

NOW I NEED:

What is the speed of the ball 0.7 seconds after it was kicked?

How high above the ground is the ball 0.7 seconds after it is kicked?

help? please.

  1. Steve

    the horizontal speed hs does not change
    the vertical speed vs is 16-9.8t
    vs(.7) = 16-9.8*.7 = 9.14

    so, the speed is sqrt(9.14^2 + 20^2) = 21.99 m/s

    vertical distance d = 1/2 at^2
    d(.7) = 4.9*.7^2 = 2.4m

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