# Physics - 2-D kinematics

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?

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1. 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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posted by Steve

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