Given:
Time taken for the ball to return to the thrower, t = 6s

To find:
(i) The velocity with which the ball was thrown up
(ii) The maximum height it reached
(iii) Its position after 4s

Solution:

1. Finding the velocity:
When a ball is thrown up vertically, its initial velocity is the same as its final velocity when it reaches the maximum height. Therefore, to find the velocity with which the ball was thrown up, we need to calculate its final velocity at the maximum height.

The formula to calculate the final velocity at maximum height is:
v = u - gt

Where:
v = final velocity (0 m/s at maximum height since the ball momentarily stops before falling back down)
u = initial velocity (unknown)
g = acceleration due to gravity (-9.8 m/s^2, negative due to the direction)
t = time taken to reach maximum height (half of the total time, 6s/2 = 3s)

Substituting the values into the formula:
0 = u - (-9.8 m/s^2)(3s)
0 = u + 29.4 m/s

Solving the equation, we find:
u = -29.4 m/s

Therefore, the velocity with which the ball was thrown up is 29.4 m/s in the opposite direction.

2. Finding the maximum height:
The maximum height can be calculated using the formula:
h = ut + (1/2)gt^2

Where:
h = maximum height (unknown)
u = initial velocity (-29.4 m/s, negative due to the direction)
g = acceleration due to gravity (-9.8 m/s^2, negative due to the direction)
t = time taken to reach maximum height (half of the total time, 6s/2 = 3s)

Substituting the values into the formula:
h = (-29.4 m/s)(3s) + (1/2)(-9.8 m/s^2)(3s)^2
h = -88.2 m - 44.1 m
h = -132.3 m

Since height cannot be negative, we take the magnitude of the value, resulting in the maximum height of 132.3 meters.

3. Finding the position after 4s:
To find the position after 4s, we need to calculate the distance traveled by the ball during this time period.

The formula to calculate the distance is:
s = ut + (1/2)gt^2

Where:
s = distance traveled (unknown)
u = initial velocity (-29.4 m/s, negative due to the direction)
g = acceleration due to gravity (-9.8 m/s^2, negative due to the direction)
t = time (4s)

Substituting the values into the formula:
s = (-29.4 m/s)(4s) + (1/2)(-9.8 m/s^2)(4s)^2
s = -117.6 m - 78.4 m
s = -196 m

Since position cannot be negative, we take the magnitude of the value, resulting in the position after 4s of

(i)29.4 m/s(ii)44.1m(iii)39.2 m from ground