Arock is projected from the edge of the top of a building with an initial velocity of 21.6 m/s at an angle of 37◦ above the horizontal. the rock strikes the ground a horizontal distance of 95 m from the base of the building. the acceleration of gravity is 9.8 m/s 2 . assume: the ground is level and that the side of the building is vertical. neglect air friction. what is the horizontal component of the rock’s velocity when it strikes the ground?

Explanation:

In horizontal direction there is no acceleration or deceleration for a rock projected at an initial angle of 37° off the ground.

So the horizontal component of velocity always remains the same.

Horizontal component of velocity is the cosine component of velocity.

Initial velocity, u = 21.6 m/s

Angle, θ = 37°

Horizontal component of velocity = u cosθ

Horizontal component of velocity = 21.6 cos37

Horizontal component of velocity = 17.25 m/s

Since the horizontal velocity is unaffected, we have

    Horizontal component of the rock’s velocity when it strikes the ground = 17.25 m/s

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

less than one you can't create energy