A soccer player kicks a soccer ball initially at rest setting it in motion at a velocity of 30 m/s. If the ball has a mass of 0.50 kg and the time of contact is 0.025 s, what is the force exerted on the player’s foot?

 F= 600 N

Explanation:

Given that

Initial velocity ,u= 0 m/s

Final velocity ,v= 30 m/s

mass ,m = 0.5 kg

time ,t= 0.025 s

The change in the linear momentum is given as

ΔP= m (v - u)

ΔP= 0.5 ( 30 - 0 ) kg.m/s

ΔP= 15 kg.m/s

We know that from second law of Newtons

Now by putting the values

F= 600 N

an energy transformation flow diagram is shown

where is it ?

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .