2. A 0.18kg baseball is thrown straight upwith a speed of 35m/s. By the time it

Question

2. A 0.18kg baseball is thrown straight upwith a speed of 35m/s. By the time it reaches the height of 38m thespeed of the ball has been reduced to 12m/s.

A.  Determine the average force of air resistance during this trip.

B.   Howmany joules of waste heat are produced as the ball comes to rest atits peak at 43.1m while slowing down from 12m/s?

C.   Ifthe ball falls to the earth from the peak height with the sameaverage force of air resistance as you found in part A, what willthe impact speed be?

    a.Force of air resistance = .796N

    b.Waste heat = 3.96J

    c.Impact speed = 21.5m/s

Explanation / Answer

m = 0.18 kg, v0 = 35 m/s, h = 38 m, v = 12 m/s a) find F initial energy Ei = mv02/2 final energy Ef = mv2/2 + mgh work by air resistance W = F*h*cos(180) = -Fh use W = Ef - Ei F = m(v02 - v2)/(2h) - mg =0.796 N b) h' = 43.1 m heat energy = Ef - mgh' = mv2/2 + mg(h - h')= 3.96 J c) find vf mgh' = mvf2/2 + F*h' vf = [2(g - F/m)h'] = 21.5 m/s d) slower, because the air resistance is always opposite to thevelocity so does negative work.

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