Answer to Question #177416 in Mechanics | Relativity for Grace Wong

Question

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?

Answer 1

Solution.

$m=0.28kg;$

$h=12m;$

$v-?;$

$mgh=\dfrac{mv^2}{2}\implies v=\sqrt{2gh}$ ;

$v=\sqrt{2\sdot9.8ms^{-2}\sdot12m}=15.34ms^{-1};$

Answer: $v=15.34ms^{-1}.$