Question

A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes airborne. The ramp makes an angle of 22.0º to the ground, and the ball travels a distance of 5.00 m on the ramp. What is the maximum height the ball reaches, above the point where it was kicked, if (a) the ramp is frictionless and (b) there is a coefficient of friction of 0.150 between the ramp and the ball? Assume the ball slides, rather than rolls, up the ramp.

Answer 1

Answer:

Explanation:

given,

initial velocity of the ball = 20 m/s

angle of ramp = 22°

ball travel at a distance  = 5 m

a) for friction less  

  $\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh$

             $v^2 = u^2 - 2gh$

             $v = \sqrt{u^2- 2 g h cos 22^0}$

             $v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0 }$

                    v = 17.58 m/s

b) considering the friction

  $\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh-\mu_kmgl$

             $v^2 = u^2 - 2gh-2\mu_kmgl$

             $v = \sqrt{u^2- 2 g h cos 22^0-2\mu_kgl}$

             $v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0-2\times 0.15\times 9.8 \times 5 }$

                   v = 17.16 m/s