Final answer:
Using energy conservation and kinematic equations, it was determined that the maximum height a ball attains above its launch point, when its speed at 5.20 m is one-half its launch speed, is 10.40 meters.
Step-by-step explanation:
The question involves the physics concept of projectile motion, specifically dealing with a ball thrown vertically upward. According to the conservation of energy, the kinetic energy at the launch will be equal to the potential energy at the maximum height, where the ball has zero kinetic energy. Taking into account that at 5.20 meters the ball has one-half of its launch speed, we can utilize the kinematic equations to derive the ball's maximum height above its launch point. The initial kinetic energy (KEinitial) at launch must be equal to the sum of the kinetic energy (KE5.20m) at 5.20 meters and the potential energy (PE5.20m) at that height:
KEinitial = KE5.20m + PE5.20m
Since KE is proportional to the square of the velocity (v), if the speed at 5.20 meters is one-half the launch speed (v/2), then the kinetic energy at that point is one-fourth the initial kinetic energy. Let 'm' be the mass of the ball and 'g' the acceleration due to gravity:
(1/2)*m*v2 = (1/4)*m*v2 + m*g*5.20
Solving for 'v', we find:
v2 = 4*g*5.20
The total energy at the maximum height is equal to the initial kinetic energy:
PEmax = (1/2)*m*v2
But PEmax = m*g*hmax, where hmax is the maximum height. So:
m*g*hmax = (1/2)*m*4*g*5.20
We can cancel out 'm' and 'g' to solve for hmax:
hmax = 2*5.20 = 10.40 meters
Therefore, the maximum height the ball attains above its launch point is 10.40 meters.