It should be Constant speed. The line goes straight & doesn’t change within the graph.
Here, Initial momentum = mu = 6*2 = 12 Kg m/s
Final momentum = mv = 6*4 = 24 Kg m/s
In short, Your Answer would be Option C
Hope this helps!
The correct answer is D. I alread took this test.
As ball is projected up in air at an angle of 45 degree without any air resistance
Let the initial speed will be v
now we will have
In x direction

in y direction

now displacement in x direction

displacement in y direction

now from above two equations we have


so above equation is a quadratic equation and hence it will be a parabolic curve
so correct answer will be
<em>C. parabolic curve.</em>