Askateboarder shoots off a ramp with a velocity of 7.1 m/s, directed at an angle of 61° above the horizontal. the end of the ramp is 1.4 m above the ground. let the x axis be parallel to the ground, the y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) how high above the ground is the highest point that the skateboarder reaches?

Highest point reached  = 3.37 m

Explanation:

Initial velocity, = 7.1 m/s

Initial vertical velocity = 7.1 sin 61 = 6.21 m/s

Consider the vertical motion of skateboarder,

We have equation of motion, v² = u² + 2as

          Initial velocity, u = 6.21 m/s

          Acceleration, a = -9.81 m/s²

          Final velocity, v = 0 m/s

         Substituting

                       v² = u² + 2as

                       0² = 6.21² + 2 x -9.81 x s

                       s = 1.97 m

So from ramp the position it goes up by 1.97 m

       Highest point reached = 1.97 + 1.4 = 3.37 m    

it breaks and dirupture other materials and other sencews from the bottom