Aphysics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. she throws the softball with a velocity of 21.5 m/s at an angle of 33.5∘ above the horizontal. when the softball leaves her hand, it is 13.5 m above the water. how far does the softball travel horizontally before it hits the water?

58.5 m

Explanation:

First of all, we need to find the total time the ball takes to reach the water. This can be done by looking at the vertical motion only.

The initial vertical velocity of the ball is

where

u = 21.5 m/s is the initial speed

is the angle

Substituting,

The vertical position of the ball at time t is given by

where

h = 13.5 m is the initial heigth

is the acceleration of gravity (negative sign because it points downward)

The ball reaches the water when y = 0, so

Which gives two solutions: t = 3.27 s and t = -0.84 s. We discard the negative solution since it is meaningless.

The horizontal velocity of the ball is

And since the motion along the horizontal direction is a uniform motion, we can find the horizontal distance travelled by the ball as follows:

a and c

c is 100% but i am confused in a

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 .