Initial position is 1.0 m Initial velocity is 6.3 m/s, up a basketball is tossed

Question

Initial position is 1.0 m
Initial velocity is 6.3 m/s, up

a basketball is tossed straight up, you watch the ball rise and then drop back to the ground

What is the magnitude and direction of the acceleration as the ball goes up?
What is the magnitude and direction of the acceleration as the ball goes down?
The problem is: What is the max height about ground reached by the ball and what are the magnitude and direction of the velocity of the ball just before it hits the ground?

I know its a kinematics problem and I have tried messing around with it but haven't been successful

Explanation / Answer

The magnitude of the deceleration = 9.8 m/sec^2,it is downward in direction

The magnitude of the acceleration = 9.8 m/sec^2 , it is downward in direction

At maximum height Final velocity , v = 0 m/sec

Thus apply first equation of motion

v^2 - u^2 = -2gh

Thus h = u^2/2g = 2.025 m

 

Velocity of the ball when it hits the ground = 6.3 m/sec in the downward direction.

 

This is because th is no loss of Kinetic energy.

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