A 0.10 kg ball of dough is thrown straight up into the air with an initial speed of 15 m/s.

1 answer:

6 0

<span>Mass of the ball is m = 0.10kg Initial speed of the Ball v = 15m/s a. When the ball is at maximum height the velocity is 0 Momentum of ball = mass x velocity Momentum = 0.10kg x 0 = 0 b. Getting the maximum height, Using the conservation of energy equation KEinitial = mgh 1/2mVin^2 = mgh => h = v^2/2g h = 15^2/2x9.8 = 11.48m => Half Height h = 5.96m Applying the conservation of energy equation at halfway V^2 = 2gh V = square root of (2x9.8x5.96) => V = square root of (116.816) So the velocity at the half way V = 10.81 m/s Momentum M = m x V => M = 0.10 x 10.81 => M = 1.081kg-m/s</span>

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Nutka1998 [239]

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