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>
With acceleration

and initial velocity

the velocity at time <em>t</em> (b) is given by




We can get the position at time <em>t</em> (a) by integrating the velocity:

The particle starts at the origin, so
.



Get the coordinates at <em>t</em> = 8.00 s by evaluating
at this time:


so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).
Get the speed at <em>t</em> = 8.00 s by evaluating
at the same time:


This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:

Answer:
Element Is The Answer I think
The rope will remain taut until the particle makes 79⁰ angle.
<h3>Change in kinetic energy of the particle</h3>
The change in kinetic energy of the particle is calculated as follows;
ΔK.E = K.Ei - K.Ef
Before the particle will achieve the given angular displacement, it will touch two new corners. Total kinetic energy lost = 30%
ΔK.E = 100%K.E - 30%K.E = 70%K.E = 0.7K.E
- let the vertical displacement of the particle = h
- horizontal length = side of the prism = a
- hypotenuse side = length of the pendulum = L
<h3>Apply principle of conservation of energy</h3>
K.E = P.E
0.7K.E = mgh
0.7(¹/₂mv²) = mg(Lsinθ)
0.7(v²) = 2g(Lsinθ)
from third kinematic equation;
v² = u² + 2gh
v² = 0 + 2gh
v² = 2g(a tanθ)
0.7(2g(a tanθ)) = 2g(Lsinθ)
0.7(a tanθ) = Lsinθ
0.7a/L = sinθ/tanθ
0.7a/L = cosθ
(0.7 x 0.8)/(3) = cosθ
0.1867 = cosθ
θ = cos⁻¹(0.1867)
θ = 79⁰
Thus, the rope will remain taut until the particle makes 79⁰ angle.
Learn more about kinetic energy here: brainly.com/question/25959744
#SPJ1