What is the initial velocity of a rock thrown vertically that reaches a maximum height of 15 m?

To determine the initial velocity of a rock thrown vertically that reaches a maximum height of 15 meters, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement. The specific equation we use is:

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

where:

• ( v ) is the final velocity (0 m/s at the maximum height),

• ( u ) is the initial velocity,

• ( a ) is the acceleration (which is -9.8 m/s² due to gravity, acting downward),

• ( s ) is the displacement (15 m in this case).

At the maximum height, the final velocity ( v ) is 0 m/s. Therefore, rearranging the equation gives us:

[ 0 = u^2 + 2(-9.8)(15) ]

This simplifies to:

[ u^2 = 2 \times 9.8 \times 15 ]

Calculating the right side:

• First, calculate ( 2 \times 9.8 = 19.6 ).

• Next, multiply by 15 to get ( 19.6 \times 15 = 294 ).