1. You throw a ball vertically upward fromthe roof of a tall building. The ball
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Question
1. You throw a ball vertically upward fromthe roof of a tall building. The ball leaves your hand at apoint even with the roof’s railing with an upward speed of16.9 m/s. On its way back down, the ball just misses therailing. Determine:
(A) the position and velocity of the ball 1.5 and3.5 seconds after leaving your hand,
(B) the velocity of the ball when it is 4.8 mabove the railing (both on the way up AND on the way down),
(C) the maximum height reached by the balland the time at which this maximum height is reached, and
(D) the acceleration of the ball when it is atits maximum height.
Explanation / Answer
A) the velocity with time is:(let upward is positive) v(t) = v(0) - gt = 16.9m/s - 9.8m/s2 * t v(1.5s) = 16.9m/s - 9.8m/s2 * 1.5s = 2.2 m/s (it is upward) v(3.5s) = 16.9m/s - 9.8m/s2 * 3.5s = -17.4m/s (it is downward) B) the heghest point is h = v(0)2 / 2g =(16.9m/s)2 / (2*9.8m/s2) = 14.57m let the speed above 4.8m above the railing is v, for energyconservation, we have: (1/2)mv(0)2 = mgh+(1/2)mv2 v = [v(0)2 - 2gh] =[(16.9m/s)2 - 2*9.8m/s2*4.8m] = 13.84m/s the velocity of the ball when it is 4.8 m above therailing (both on the way up AND on the way down), is 13.84 m/s C) the heghest point is h = v(0)2 / 2g =(16.9m/s)2 / (2*9.8m/s2) = 14.57m the time is: t = v(0) / g = 16.9m/s / 9.8m/s2 =1.74s D) the acceleration at max height is g =-9.8m/s2 (upward is positive) . v = [v(0)2 - 2gh] =[(16.9m/s)2 - 2*9.8m/s2*4.8m] = 13.84m/s the velocity of the ball when it is 4.8 m above therailing (both on the way up AND on the way down), is 13.84 m/s C) the heghest point is h = v(0)2 / 2g =(16.9m/s)2 / (2*9.8m/s2) = 14.57m the time is: t = v(0) / g = 16.9m/s / 9.8m/s2 =1.74s the heghest point is h = v(0)2 / 2g =(16.9m/s)2 / (2*9.8m/s2) = 14.57m the time is: t = v(0) / g = 16.9m/s / 9.8m/s2 =1.74s D) the acceleration at max height is g =-9.8m/s2 (upward is positive) .Related Questions
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