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?

Answer 1

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball,
`v_(o) = 20.0 m/s`

Angle made by the ramp,
`\theta = 22.0^(\circ)`

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:

`v^(2) = v_(o)^(2) - 2gH`

where

H =
`dsin22^(\circ) = 5sin22^(\circ)`

g =
`9.8 m/s^(2)`

`v^(2) = 20^(2) - 2* 9.8* 5sin22^(\circ)`

`v = \sqrt{400 - 19.6* 5sin22^(\circ)}` = 19.06 m/s

Now, maximum height attained is given by:

`h = ((vsin\theta)^(2))/(2g)`

`h = ((19sin(22^(\circ)))^(2))/(2* 9.8) = 2.60 m`

Height from the ground =
`5sin22^(circ) + 2.86 = 1.87 + 2.60 = 4.47m`

(b) now, considering the coefficient of friction bhetween ramp and the ball,
`\mu = 0.150`:

velocity can be given by the eqn-3 of motion:

`v^(2) = v_(o)^(2) - 2gH - \mu gd`

`v^(2) = 20^(2) - 2* 9.8* 5sin22^(\circ) - 0.150* 9.8* 5`

`v = \sqrt{400 - 19.6* 5sin22^(\circ) - 0.150* 9.8* 5}` = 18.7 m/s

Now, maximum height attained is given by:

`h = ((vsin\theta)^(2))/(2g)`

`h = ((18.7sin(22^(\circ)))^(2))/(2* 9.8) = 2.50 m`

Height from the ground =
`5sin22^(circ) + 2.86 = 1.87 + 2.50 = 4.37 m`