A ball is thrown vertically upward. The quantity which stays constant is a) displacement b)acceleration c)velocity d)speed e)momentum.
2) A ball is thrown from the ground into the air. At a height of 9.6m, the velocity is observed to be v= 8.0i + 6.5j in m/s. Find the maximum height the ball can reach.
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Expert's answer
2017-07-06T10:28:07-0400
The acceleration stays constant or the answer b)
The second part: Horizontal component of velocity is 8m/s The velocity at 9.6m v = sqrt(8^2 + 6.5^2) v = 10.3 m/s The whole energy at 9.6m is E = 0.5mv^2 + mgh E = m(0.5*10.3^2 + 9.8*9.6) E = 147.125m Energy at initial position is equal to the Energy at final position 147.125m = mgh + 0.5mv^2 Maximum height v = horizontal compnent of velocity 147.125m = m*9.8*H + 0.5*m*64 147.125 = 9.8H + 32 H = 11.8 m
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