It would be measured in newtons :)
The tangent looks good.
The curve is a bit crooked, at the 0.9 and 1.
But overall, cool graph.
Answer:
Part a)

Part b)

Part c)

Part d)

Part e)

Part f)

Part g)

Explanation:
Initial speed of the launch is given as
initial speed = 
angle =
degree
Now the two components of the velocity

similarly we have

Part a)
Now we know that horizontal range is given as

maximum height is given as

so we have

time of flight is given as



Part b)
Now the speed of the ball in x direction is always constant
so at the peak of its path the speed of the ball is given as



Part c)
Initial vertical velocity is given as


Part d)
Initial speed is given as

so we will have


Part e)
Angle of projection is given as



Part f)
If we throw at same speed so that it reach maximum height
then the height will be given as


Part g)
For maximum range the angle should be 45 degree
so maximum range is


Answer:
The horizontal force is 106.89 N.
Explanation:
Given that,
Work done = 310 J
Distance = 2.9 m
We need to calculate the horizontal force
Using formula of work done

Where, 

Put the value into the formula



Hence, The horizontal force is 106.89 N.
Explanation:
Total surface area of cylinder:

sheet of metal required = 120.95 cm^2