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March 29, 2015

March 29, 2015

Posted by **Anonymous** on Friday, February 3, 2012 at 4:37pm.

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.

- Physics - 2-D kinematics -
**Steve**, Friday, February 3, 2012 at 5:01pmthe 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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