Respuesta :
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
[tex]\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh[/tex]
[tex]v^2 = u^2 - 2gh[/tex]
[tex]v = \sqrt{u^2- 2 g h cos 22^0}[/tex]
[tex]v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0 }[/tex]
v = 17.58 m/s
b) considering the friction
[tex]\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh-\mu_kmgl[/tex]
[tex]v^2 = u^2 - 2gh-2\mu_kmgl[/tex]
[tex]v = \sqrt{u^2- 2 g h cos 22^0-2\mu_kgl}[/tex]
[tex]v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0-2\times 0.15\times 9.8 \times 5 }[/tex]
v = 17.16 m/s
