Given information:
- The object is thrown vertically upwards.
- The object returns to the thrower after 6 seconds.

Assumptions:
- We are neglecting air resistance.
- The acceleration due to gravity is constant and equal to 9.8 m/s².

Calculating the initial velocity:
To calculate the initial velocity at which the object was thrown upwards, we can use the equation of motion:

v = u + at

Where:
v = final velocity (0 m/s, as the object reaches its highest point and then starts to fall back down)
u = initial velocity (to be determined)
a = acceleration due to gravity (-9.8 m/s², as it acts in the opposite direction to the motion)
t = time taken (6 seconds)

0 = u + (-9.8 × 6)
u = 58.8 m/s

Therefore, the object was thrown upwards with an initial velocity of 58.8 m/s.

Calculating the maximum height:
To calculate the maximum height reached by the object, we can use the equation of motion:

s = ut + (1/2)at²

Where:
s = displacement (to be determined)
u = initial velocity (58.8 m/s)
t = time taken to reach maximum height (half of the total time, i.e., 6/2 = 3 seconds)
a = acceleration due to gravity (-9.8 m/s²)

s = (58.8 × 3) + (1/2)(-9.8)(3²)
s = 176.4 - 44.1
s = 132.3 m

Therefore, the object rises to a maximum height of 132.3 meters.

Plotting the speed-time graph:
To plot the speed-time graph for the object, we need to consider the following:

- At the start, the object has an initial velocity of 58.8 m/s and is moving upwards.
- The speed decreases uniformly due to the acceleration due to gravity until it reaches zero at the maximum height.
- After reaching the maximum height, the speed increases uniformly in the downward direction due to gravity.

The speed-time graph would look like this:

```
Speed (m/s)
| +
| + +
| + +
|+ +
+------------------- Time (s)
0 3 6
```

In the graph, the speed initially increases in the positive direction (upwards) and then decreases until it reaches zero at 3 seconds (maximum height). After that, the speed increases in the negative direction (downwards) until it reaches the initial velocity of -58.8 m/s at 6 seconds (when it returns to the thrower).

This graph represents the change in speed of the object over time during its upward and downward motion.

the upwards velocity will be 30 m/s .to plot the graph. we have acceleration due to gravity constant. and its value is ~10ms^-2. now when the ball is thrown up wards the direction of acceleration due to gravity and velocity is opposite, so the velocity will decrease but there will be a linear decrease in velocity as v=u+at is equivalent to y=c+mx for velocity time graph. and m=-10 (slope of the line basically). There will be a point when v=0 . the graph at that point will change as now the velocity and acceleration due to gravity will be in same direction and slope of the straight line m=10 positive slope. The velocity will then increase to its original value just before hitting the ground. Good luck