Answer to Question #3112 in Electric Circuits for Jennifer

Question #3112
A juggler throws a ball with an initial horizontal velocity of +1.1 m/s and an initial vertical velocity of +5.7 m/s. What is its acceleration at the top of its flight path? Make sure to consider the sign when responding. Consider the upward direction as positive.
1
Expert's answer
2011-06-15T11:56:28-0400
Lets write down law of motion.
x=V0x t Vx=V0x ax=0
y=V0yt - g t2/2 Vy=V0y-gt ay=-g |a| = √(ax2+ay2) = √(0+(-g)2)=g

where g is acceleration relative to free-fall.

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