Respuesta :
Answer:
(a) Height is 4.47 m
(b) Height is 4.37 m
Solution:
As per the question:
Initial velocity of teh ball, [tex]v_{o} = 20.0 m/s[/tex]
Angle made by the ramp, [tex]\theta = 22.0^{\circ}[/tex]
Distance traveled by the ball on the ramp, d = 5.00 m
Now,
(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:
[tex]v^{2} = v_{o}^{2} - 2gH[/tex]
where
H = [tex]dsin22^{\circ} = 5sin22^{\circ}[/tex]
g = [tex]9.8 m/s^{2}[/tex]
[tex]v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}[/tex]
[tex]v = \sqrt{400 - 19.6\times 5sin22^{\circ}}[/tex] = 19.06 m/s
Now, maximum height attained is given by:
[tex]h = \frac{(vsin\theta)^{2}}{2g}[/tex]
[tex]h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m[/tex]
Height from the ground = [tex]5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m[/tex]
(b) now, considering the coefficient of friction bhetween ramp and the ball, [tex]\mu = 0.150[/tex]:
velocity can be given by the eqn-3 of motion:
[tex]v^{2} = v_{o}^{2} - 2gH - \mu gd[/tex]
[tex]v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5[/tex]
[tex]v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5}[/tex] = 18.7 m/s
Now, maximum height attained is given by:
[tex]h = \frac{(vsin\theta)^{2}}{2g}[/tex]
[tex]h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m[/tex]
Height from the ground = [tex]5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m[/tex]
