Perpendicularly 81951

A body of mass 4g is shot perpendicularly from the ground and lands in 8s. What is its kinetic energy at the beginning of its trajectory, what speed does it take off, and where does the distance from the ground stop? Thank you

Correct answer:

E1 =  3.0796 J
v =  39.24 m/s
s =  78.48 m

Step-by-step explanation:

m=4 g kg=4:1000  kg=0.004 kg t=8 s  t1=t/2=8/2=4 s g=9.81 m/s2  s=21 g t12=21 9.81 42=251962=78.48 m  E1=E2  E1=m g s=0.004 9.81 78.48=3.0796 J
v=g t1=9.81 4=39.24 m/s

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