2. + 41 points M14 6.8.056. DMy Notes You throw a ball of mass 0.2 kg straight u

Question

2. + 41 points M14 6.8.056. DMy Notes You throw a ball of mass 0.2 kg straight up. You observe that it takes 3.5 s to go up and down, returning to your hand. Assuming we can neglect air resistance, the time it takes to go up to the top is half the total time, 1.75s. Note that at the top the momentum is momentarily zero, as it changes from heading upward to heading downward. (a) Use the momentum principle to determine the speed that the ball had just AFTER it left your hand. Vinitial = m/s (b) Use the Energy Principle to determine the maximum height above your hand reached by the ball. h = m Additional Materials V section 6.8

Explanation / Answer

(a)

p(top) = momentum at top =0

F = deltaP/t
- mg = [p(top) - p(hand))/1.75
- 2*9.8 = - p(hand)/1.75

vin = 9.8*1.75

vin = 17.15 m/s

(b)

energy in hand = energy at top height
KE + PE = KE1+PE1
0.5*m*vin2 + mg*0 = 0 (top rest) + m*g*H
H = max height = vin2/2g

H = 17.152/2*9.8
H = 15 m

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