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.
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?
Physics - 2-D kinematics -
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