# A bb player crouches lowering center of gravity 0.3 m. when he jumps, center of gravity reaches...

## Question:

A bb player crouches lowering center of gravity 0.3 m. when he jumps, center of gravity reaches 0.9 m above normal.

A. Calculate velocity when feet left ground.

B. Calculate acceleration produced to achieve velocity.

C. What force exerted on floor if mass is 110 kg?

## Conservation Of Energy:

The total amount of energy contained within the system remains conserved during the conversion process it can be converted from one form into another for example potential energy contained within the system can be converted into the kinetic energy when the object is allowed to fall.

Given

The distance by which the centre of gravity is lowered is {eq}s_1= 0.3\ m {/eq}

The distance of gravity above normal is {eq}s_2= 0.9\ m {/eq}

When the feet left the ground then the velocity can find out by using the conservation of energy principle according to which the potential energy at the highest point will be equal to the kinetic energy:

{eq}mgs_2=\frac{1}{2}mv^2\\ v=\sqrt{2gs_2}\\ v=\sqrt{2\times 9.81\times 0.9 }\\ v=4.20\ m/s {/eq}

Thus, the velocity by which the feet left the ground is 4.20 m/s

(b)

Now for the acceleration produced to achieve velocity:

{eq}v^2=u^2+2as_1\\ (4.20)^2=0+2\times a\times 0.3\\ a=29.4\ m/s^2 {/eq}

Thus, the acceleration produced to achieve velocity is 29.4 m/s^2

(c)

Now for the force exerted on the floor Work done will be equal to kinetic energy and the potential energy:

{eq}F\times 0.3=\frac{1}{2}(110)\times (4.20)^2+(110)\times 0.3\\ F=3344\ N {/eq}

Thus, the force exerted on the floor is 3344 N