A boy throws a ball straight up into the air. It reaches the highest point of its flight after 4 seconds. How fast was the ball going when it left the boy's hand?

Take into account the the Earths gravity is 9.8 meters a second

Explanation:

Gravity pulls down on the ball at g=-9.81 m/s^2. Up is positive, down is negative.The ball started at a certain initial velocity of Vi m/s. Time it took is t=4s. Final velocity is Vf=0 m/s, because at the highest point the ball stops moving.

Vf=(g*t)+Vi

Rearrange for Vi.

Vi=Vf-(g*t)

Vi=0-(-9.81*4)=39.24 m/s (upward)

Think about it this way for the non-mathematical approach. The ball stops at the top. The initial velocity gets reduced by 9.81 m/s every second, and reaches 0 m/s at the top. It took 4 seconds, so 9.81*4 is equal to the initial upward velocity.

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what type of charge does it have? negative