Aperson 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?

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball,

Angle made by the ramp,

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:

where

H =

g =

= 19.06 m/s

Now, maximum height attained is given by:

Height from the ground =

(b) now, considering the coefficient of friction bhetween ramp and the ball, :

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

= 18.7 m/s

Now, maximum height attained is given by:

Height from the ground =

force tends to be low and long diustance masses are high

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s