Asked by Jordan
A .5kg ball is dropped from a 30m high building is on the ground for 2ms then bounces up 20m. What avg force was exerted on the ball?
Answers
Answered by
Elena
The law of conservation of energy
m•g•h1=m•v1²/2
v1=sqrt(2•g•h1)=sqrt(2•9.8•30) =24.3 m/s
m•v2²/2=m•g•h2
v2=sqrt(2•g•h2)=sqrt(2•9.8•20) =19.8 m/s
The law of conservation of linear momentum
p=m•v2-m• (-v1)=m(v2+v1)=FΔt
F= m(v2+v1)/ Δt=0.5(24.3+19.8)/0.002=44100 N
m•g•h1=m•v1²/2
v1=sqrt(2•g•h1)=sqrt(2•9.8•30) =24.3 m/s
m•v2²/2=m•g•h2
v2=sqrt(2•g•h2)=sqrt(2•9.8•20) =19.8 m/s
The law of conservation of linear momentum
p=m•v2-m• (-v1)=m(v2+v1)=FΔt
F= m(v2+v1)/ Δt=0.5(24.3+19.8)/0.002=44100 N
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.