Abasketball player standing under the hoop launches the ball straight up with an initial velocity of v₀ = 3.25 m/s from 2.5 m above the ground.
what is the maximum height, h (in meters), above the launch point the basketball will achieve?

0.53m

Explanation:

According to the equation of motion v²= v₀²+2as

Since the body is launched upward, the final velocity at the maximum height will be "zero" since the body will momentarily be at rest at the maximum height i.e v = 0

Initial velocity given (v₀) = 3.25 m/s

The body is also under the influence of gravity but the acceleration due to gravity will be negative being an upward force (a = -g) and the distance (s) will serve as our maximum height (h)

The equation of motion will.now become

V = v₀² -2gh

Where v = 0 v₀ = 3.25m/s g = 10m/s h = ?

0 = 3.25² - 2(10)h

0 = 10.56 - 20h

-10.56 = -20h

h = 10.56/20

h = 0.53m

Therefore, the maximum height, h (in meters), above the launch point that the basketball will achieve is 0.53m

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