Answer to Question #3112 in Electric Circuits for Jennifer

Question

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.

Answer 1

Lets write down law of motion.
x=V_0x t V_x=V_0x a_x=0
y=V_0yt - g t²/2 V_y=V_0y-gt a_y=-g |a| = √(a_x²+a_y²) = √(0+(-g)²)=g

where g is acceleration relative to free-fall.