Aball is thrown straight up with enough speed so that it is in the air for several seconds. assume the positive direction is upwards. part apart complete what is the velocity of the ball when it reaches its highest point? express your answer to two significant figures and include the appropriate units. v = 0 ms previous answers correct part b what is its velocity 0.70 s before it reaches its highest point?

a) v= 0 m/s b) v= 6.86 m/s

Explanation:

a) When the ball reaches to its highest point, under the influence of gravity, before starting to fall down, it momentarily comes to an stop (this is needed prior to change direction in any movement), so, applying the definition of acceleration, and replacing the acceleration a by g, we have:

vf = v₀ - g*t (1)

The minus sign means that the acceleration due to gravity is always downward, so if we assume that the positive direction is upwards it must be negative.

At the highest point, vf= 0.

b) Prior to solve this point, we need to know which is the time when the ball reaches to its highest point.

As we know vf=0, we can solve (1) for t, as follows:

th = v₀/g

Now, for a time that is 0.7 s before this time, applying the acceleration definition and solving for v again, we have:

v = v₀ -(g *(th-0.7 s)), but th= v₀/g, so we get:

v= v₀ -g((v₀/g)-0.7 s) = v₀ - v₀ + g*0.7 s

⇒ v=g*0.7 s = 9.8 m/s²*0.7 s

⇒ v = 6.86 m/s

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 .