A volleyball of mass 0.28 kg is dropped from the top of the bleachers (height equals 12 m). What is the speed of the volleyball right before it hits the gym floor?
Solution.
m=0.28kg;m=0.28kg;m=0.28kg;
h=12m;h=12m;h=12m;
v−?;v-?;v−?;
mgh=mv22 ⟹ v=2ghmgh=\dfrac{mv^2}{2}\implies v=\sqrt{2gh}mgh=2mv2⟹v=2gh ;
v=2⋅9.8ms−2⋅12m=15.34ms−1;v=\sqrt{2\sdot9.8ms^{-2}\sdot12m}=15.34ms^{-1};v=2⋅9.8ms−2⋅12m=15.34ms−1;
Answer: v=15.34ms−1.v=15.34ms^{-1}.v=15.34ms−1.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments