Answer: 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
