A ball of mass 0.16 kg is located 2.07 m above the surface of the ground. It is

Question

A ball of mass 0.16 kg is located 2.07m above the surface of the ground. It is released fromrest, falls straight down, and hits the ground.
How much work was done by the force of gravity?
I work this out to be 3.25 J however i do not understand : What is the speed of the ball just before it hitsthe ground? Neglect the effect of air friction. and What height would the ball have to be releasedfrom in order to reach a speed of 18.77 m/s(i.e., 42.00 mi/hr) just before hitting theground? A ball of mass 0.16 kg is located 2.07m above the surface of the ground. It is released fromrest, falls straight down, and hits the ground.
How much work was done by the force of gravity?
I work this out to be 3.25 J however i do not understand : What is the speed of the ball just before it hitsthe ground? Neglect the effect of air friction. and What height would the ball have to be releasedfrom in order to reach a speed of 18.77 m/s(i.e., 42.00 mi/hr) just before hitting theground? however i do not understand : What is the speed of the ball just before it hitsthe ground? Neglect the effect of air friction. and What height would the ball have to be releasedfrom in order to reach a speed of 18.77 m/s(i.e., 42.00 mi/hr) just before hitting theground?

Explanation / Answer

a.) Gravitational potential energy becomes fully converted tokinetic energy by the time it reaches the ground:      mgh =1/2mv2      2gh= v2      v = (2gh)= (2(9.8)(2.07)) ˜ 6.39 m/s b.) Now do the same thing, only this time solve for h and plugin for v:      mgh =1/2mv2      h= v2/(2g) = (18.77)2/(2(9.8)) ˜ 17.98m

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