How fast must a baseball move to reach a maximum height of 15.0 m above the point where it left your hand?

To determine how fast a baseball must move to reach a maximum height of 15.0 meters, we can use the principle of energy conservation or kinematic equations. In this case, utilizing the kinematic equation that relates initial velocity, acceleration due to gravity, and maximum height is appropriate.

The relevant equation is:

[ v^2 = u^2 + 2as ]

where:

• ( v ) is the final velocity at the maximum height (which is 0 m/s, since the baseball stops rising at the peak),

• ( u ) is the initial velocity (the speed at which the baseball is thrown),

• ( a ) is the acceleration (which is -9.81 m/s² due to gravity acting downwards),

• ( s ) is the maximum height reached (15.0 m).

Rearranging the equation to solve for ( u ):

[ 0 = u^2 + 2(-9.81)(15.0) ]

This simplifies to:

[ u^2 = 2(9.81)(15.0) ]

[ u^2 = 294.3 ]

Taking the square root gives:

[ u = \sqrt{294.3