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

c ) the 4 feet ramp

Step-by-step explanation:

got it right

Answer 2

Step-by-step 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

`(1)/(2)mv^2 = (1)/(2)mu^2 - mgh`

`v^2 = u^2 - 2gh`

`v = √(u^2- 2 g h cos 22^0)`

`v = √(20^2- 2* 9.8 * 5 cos 22^0 )`

v = 17.58 m/s

b) considering the friction

`(1)/(2)mv^2 = (1)/(2)mu^2 - mgh-\mu_kmgl`

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

`v = √(u^2- 2 g h cos 22^0-2\mu_kgl)`

`v = √(20^2- 2* 9.8 * 5 cos 22^0-2* 0.15* 9.8 * 5 )`

v = 17.16 m/s