Question #96928
Ball landed on the ground with impact velocity of 30 m/s.

a.) from how high it was dropped?
b.) where was it at time= 0.5s?
c.) how long did it take to fall?
1
Expert's answer
2019-10-23T09:53:36-0400

The final velocity of the ball is vf=30m/s,v_f=30\:\rm m/s, the acceleration due gravity g=9.8m/s2.g=9.8\:\rm m/s^2.

(a) Height


H=vf22g=3022×9.8=46mH=\frac{v_f^2}{2g}=\frac{30^2}{2\times 9.8}=46\:\rm m

(b) Position at 0.5s


y=Hgt22=469.8×0.522=44.8m(above  the  ground)y=H-\frac{gt^2}{2}=46-\frac{9.8\times 0.5^2}{2}=44.8\:\rm m\quad (above \; the \; ground)

(c) Time of motion


t=vfg=309.8=3st=\frac{v_f}{g}=\frac{30}{9.8}=3\:\rm s


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