(a) The time taken for the projectile to reach the maximum height is 32.65 s.
(b) The horizontal range of the projectile is 9,049.1 m.
The given parameters:
- Initial velocity of the projectile, u = 320 m/s
- Angle of projection, = 30 degrees
The time taken for the projectile to reach the maximum height is calculated as follows;

The horizontal range of the projectile is calculated as follows;

Learn more about horizontal range here: brainly.com/question/12870645
Answer:
⇒ 401.6 ± 7.1 ms
⇒ 57.2 ± 1.6 kg
⇒ 8.1 × 107 ± 2.9 × 105 m
⇒ 2.1 ± 8.8 µN
⇒ 294 ± 0.005 cm
Explanation:
As given all equation we do without calculation we will get here round these calculations which is given below as
a) 401.6473 ± 7.0912 ms
⇒ 401.6 ± 7.1 ms
b) 57.212 ± 1.612 kg
⇒ 57.2 ± 1.6 kg
c) 8.12754×107 ± 2.847×105 m
⇒ 8.1 × 107 ± 2.9 × 105 m
d) 2.94 ± 8.803 µN
⇒ 2.1 ± 8.8 µN
e) 294 ± 0.00481 cm
⇒ 294 ± 0.005 cm
<span>Assume: neglect of the collar dimensions.
Ď_h=(P*r)/t=(5*125)/8=78.125 MPa ,Ď_a=Ď_h/2=39 MPa
τ=(S*Q)/(I*b)=(40*〖10〗^3*π(〖0.125〗^2-〖0.117〗^2 )*121*〖10〗^(-3))/(π/2 (〖0.125〗^4-〖0.117〗^4 )*8*〖10〗^(-3) )=41.277 MPa
@ Point K:
Ď_z=(+M*c)/I=(40*0.6*121*〖10〗^(-3))/(8.914*〖10〗^(-5) )=32.6 MPa
Using Mohr Circle:
Ď_max=(Ď_h+Ď_a)/2+âš(Ď„^2+((Ď_h-Ď_a)/2)^2 )
Ď_max=104.2 MPa, Ď„_max=45.62 MPa</span>
In order to solve the problem it is necessary to take into account the concepts related to the moment of inertia, and the center of mass of the object.
Our values are given by:
m = 0.514kg
I = 4.7m (each side)
A) For the case when the axis passes through the midpoints of opposite sidesand lies in the plane of the square. The formula is given by,

B) For the case when the axis passes through the midpoint of one of the sides and is perpendicular to the plane of the square. The formula is given by,

C) For the case when the axis lies in the plane of square f passes through two diagonally opposite particles,


