a ball rolling on oa inclined plane dors so with constant acceleration. One ball, A, is released from rest on th etop of an inclined plane 18m long to reach the bottom 3s later. At the same time as A is released, another ball, B, is projected up the same plane with a certain initial velocity. B is to travel part way up the plane, stop, and return to the bottom so as to arrive there simultaneously with A. calculate the velocity of projection of B.

Question

a ball rolling on oa inclined plane dors so with constant acceleration. One ball, A, is released from rest on th etop of an inclined plane 18m long to reach the bottom 3s later. At the same time as A is released, another ball, B, is projected up the same plane with a certain initial velocity. B is to travel part way up the plane, stop, and return to the bottom so as to arrive there simultaneously with A. calculate the velocity of projection of B.

Answer 1

See my answer posted later at

http://www.jiskha.com/display.cgi?id=1364108318

Answer 2

To solve this problem, we can use the kinematic equations of motion. Let's consider the motion of each ball separately:

For Ball A:
- Initial velocity (u) = 0 (since it is released from rest)
- Distance (s) = 18m
- Time (t) = 3s

Using the equation:

s = ut + (1/2)at^2

We can rearrange the equation to solve for acceleration (a):

a = (2s - ut^2) / t^2
a = (2 * 18 - 0 * 3^2) / 3^2
a = 12 m/s^2

Now, let's consider Ball B:
- We are asked to find the initial velocity (u) of Ball B.

Both Ball A and Ball B reach the bottom of the incline in the same time (3s), so the total time taken by Ball B is also 3s.

Since Ball B is projected up the incline, its acceleration (a) will be opposite to that of Ball A. Therefore, the acceleration of Ball B is -12 m/s^2.

Using the equation:

s = ut + (1/2)at^2

We need to find the distance (s) traveled by Ball B in 3 seconds. The total distance traveled by Ball B will be the sum of the distances traveled while going up and coming back down the incline.

Let's assume the distance traveled by Ball B when it stops is 'd'. The distance traveled during the first part of the motion will be 'd' and during the second part, it will be (18 - d) (since the incline is 18m long).

For the first part of the motion:
s1 = d
u1 = initial velocity of Ball B
t1 = time taken = 3s
a1 = -12 m/s^2

s1 = u1 * t1 + (1/2) * a1 * t1^2
d = u1 * 3 + (1/2) * (-12) * 3^2
d = 3u1 - 18

For the second part of the motion:
s2 = 18 - d = 18 - (3u1 - 18) = 36 - 3u1
u2 = 0 (Ball B stops)

s2 = u2 * t2 + (1/2) * a2 * t2^2
36 - 3u1 = 0 * 3 + (1/2) * (-12) * 3^2
36 - 3u1 = -18

Now, we can solve the equations to find the value of u1 (initial velocity of Ball B).

d = 3u1 - 18
36 - 3u1 = -18

Solving these two equations simultaneously will give us the value of u1.