The kicker now kicks the ball with the same speed as in the number of 4,but at 60.0°from the horizontal or 30.0° from the vertical. a.What is the time the ball is in the air? b.What is the distance the balls travel before it hits the ground? c.What is the maximum height?

Explanation:

1) The time of flight equation for projectile motion can be used here to find total time in air.

t = 2vsin∅ / g

where v is speed, Ф is launch angle

t = 2×4×sin 60 / 9.8

t = 0.71 seconds

2) Distance where it hit the ground is called as range and has the following standard equation

D = v² sin2Ф/g

D = 4²sin 2×60 / 9.8

D = 1.41m

3) Maximum elevation is maximum time reached

h = v² sin²Ф / 2g

h = 4²sin² 60 / 2*9.8

h = 0.61 m

A) the initial point all energy is elastic potential and the final point all energy is kinetic

B) a bar graph the two bars have the same height and the sum of their height is the initial energy

C) two bars, one for the  kinetic energy and the other for the gravitational potential energy.

Explanation:

A) For this exercise we must use the energy conservation relations

starting point. When the spring is compressed

        Em₀ = K_e = ½ k x²

end point, at the bottom of the loop

        Em_f = K = ½ m v²

energy is conserved

        Em₀ = Em_f

         ½ k x² = ½ m v²

         v =   x

In a bar graph the initial point all energy is elastic potential and the final point all energy is kinetic

B) intermediate point in a quarter of the radius

In this case we use the lower part of the loop as the starting point and the quarter part of the bow as the end point.

        Em₀ = K

         Em_f = K + U = ½ m v² + m g R

in a bar graph the two bars have the same height and the sum of their height is the initial energy

C) End point highest part of the loop

   starting point, bottom of loop

         Emo = K = ½ m v₀²

from part A of the exercise we saw that it is equal to the elastic energy of the spring

    final point. Highest part of the loop

         Emf = K + U

         Em_f = ½ m + mg (2R)

where R is the radius of the loop

         Em₀ = Em_f

        1/2 m v₀² = 1/2 m v_{f}^2+ mg 2R

        v₀² = v_f^2 + 4gR

In a bar graph there are two bars, one for the  kinetic energy and the other for the gravitational potential energy. The sum of the heights of these bars is the initial energy, so the energy is transformed but not created or destroyed in the process.