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.

Expert's answer

Lets write down law of motion.

x=V_{0x} t V_{x}=V_{0x} a_{x}=0

y=V_{0y}t - g t^{2}/2 V_{y}=V_{0y}-gt a_{y}=-g |**a**| = √(a_{x}^{2}+a_{y}^{2}) = √(0+(-g)^{2})=g

where g is acceleration relative to free-fall.

x=V

y=V

where g is acceleration relative to free-fall.

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