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
1) What is the maximum height above ground reached by the ball? Draw a sketch. Choose origin, coordinate direction. Show inventory list - what is known?

2) What are the magnitude and the direction of the velocity of the ball just before it hits the ground?

3) Write the kinematic equation(s) and solution of 1)

4) Write the kinematic equation(s) and solution of 2)

Explanation / Answer

1)

Using the equation of motion :

v^2 = u^2 + 2as

here , v = final velocity at the top of its trajectory = 0

u = 6.3 m/s

a = -9.8 m/s2

s = displacment

So, 0^2 = 6.3^2 + 2*(-9.8)*s

So, s = 2.03 m

So, maximum height above ground = 1 m + 2.03 m = 3.03 m

2)

Using the same equation above:

v^2 = u^2 + 2as

Now, at the bottom( ground), v = final velocity at ground

u = 6.3 m/s2

a = -9.8 m/s2

s = -1 m (displacement is negative 1m (it comes down by 1 m as it was originally 1m above ground))

So, v^2 = 6.3^2 + 2*(-9.8)*(-1)

So, v = 7.7 m/s <--------answer

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