A ball thrown up verically returns to the thrower after 6 s. Find
(a) the velocity with which it was thrown up. (b) the maximum height it reaches, and (c ) its position after 4 s.

Explanation

To solve the problem step by step, we will break down each part of the question and apply the relevant physics concepts.

### Given:
- Total time for the ball to return to the thrower, \( T = 6 \, \text{s} \)
- Time taken to reach maximum height, \( t = \frac{T}{2} = \frac{6}{2} = 3 \, \text{s} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \) (acting downwards)

### (a) Finding the initial velocity \( u \):
Using the first equation of motion:
\[ v = u + at \]
Where:
- \( v = 0 \, \text{m/s} \) (final velocity at maximum height)
- \( a = -g = -9.8 \, \text{m/s}^2 \) (acceleration is negative as it is against the motion)
- \( t = 3 \, \text{s} \)

Rearranging the equation to find \( u \):
\[ 0 = u - 9.8 \times 3 \]
\[ u = 9.8 \times 3 \]
\[ u = 29.4 \, \text{m/s} \]

### (b) Finding the maximum height \( h \):
Using the third equation of motion:
\[ v^2 = u^2 + 2as \]
Where:
- \( v = 0 \, \text{m/s} \)
- \( u = 29.4 \, \text{m/s} \)
- \( a = -9.8 \, \text{m/s}^2 \)
- \( s = h \) (maximum height)

Rearranging the equation to find \( h \):
\[ 0 = (29.4)^2 + 2 \times (-9.8) \times h \]
\[ (29.4)^2 = 19.6h \]
\[ h = \frac{(29.4)^2}{19.6} \]
Calculating \( h \):
\[ h = \frac{864.36}{19.6} \approx 44.1 \, \text{m} \]

### (c) Finding the position after 4 seconds:
After 3 seconds, the ball reaches maximum height and starts descending. The position after 4 seconds means it has been falling for 1 second.

Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( u = 0 \, \text{m/s} \) (initial velocity when falling)
- \( a = 9.8 \, \text{m/s}^2 \) (positive as it is in the direction of motion)
- \( t = 1 \, \text{s} \)

Calculating the distance fallen in 1 second:
\[ s = 0 \times 1 + \frac{1}{2} \times 9.8 \times (1)^2 \]
\[ s = \frac{1}{2} \times 9.8 \approx 4.9 \, \text{m} \]

Now, to find the position after 4 seconds:
\[ \text{Position after 4 seconds} = \text{Maximum height} - \text{Distance fallen} \]
\[ \text{Position after 4 seconds} = 44.1 \, \text{m} - 4.9 \, \text{m} = 39.2 \, \text{m} \]

### Final Answers:
(a) The velocity with which it was thrown up is \( 29.4 \, \text{m/s} \).
(b) The maximum height it reaches is \( 44.1 \, \text{m} \).
(c) Its position after 4 seconds is \( 39.2 \, \text{m} \).