I would say your answer to this question would be D
        
                    
             
        
        
        
The height of the ball above the ground is 38.45 m
First we will calculate the velocity of the ball when it touch the ground by using first equation of motion
v=u+gt
v=0+9.81×2.8
v=27.468 m/s
now the height of the ground can be calculated by the formula
v=√2gh
27.468=√2×9.81×h
h=38.45 m
 
        
             
        
        
        
Answer:
a) (0, -33, 12)
b) area of the triangle : 17.55 units of area 
Explanation:
<h2>
a) </h2>
We know that the cross product of linearly independent vectors  and
 and  gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.
 gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a. 
Luckily for us, we know that vectors  and
 and  are living in the plane through the points P, Q, and R, and are linearly independent.
 are living in the plane through the points P, Q, and R, and are linearly independent.
We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).
If they weren't linearly independent, we will obtain vector zero as the result of the cross product.
So, for our problem:







<h2>B)</h2>
We know that  and
 and  are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:
 are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:

so:




 
        
             
        
        
        
Answer:
a)  Em₀ = 42.96 104 J
, b)    = -2.49 105 J
, c)  vf = 3.75 m / s
 = -2.49 105 J
, c)  vf = 3.75 m / s
Explanation:
The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has
         Em = K + U
a) Let's look for the initial mechanical energy
       Em₀ = K + U
       Em₀ = ½ m v2 + mg and
       Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142
       Em₀ = 36 104 + 6.96 104
       Em₀ = 42.96 104 J
b) The work of the friction force is equal to the change in the mechanical energy of the body
      = Em₂ -Em₀
 = Em₂ -Em₀
      Em₂ = K + U
      Em₂ = ½ m v₂² + m g y₂
      Em₂ = ½ 50 85 2 + 50 9.8 427
      Em₂ = 180.625 + 2.09 105
      Em₂ = 1,806 105 J
       = Em₂ -Em₀
 = Em₂ -Em₀
       = 1,806 105 - 4,296 105
 = 1,806 105 - 4,296 105
       = -2.49 105 J
 = -2.49 105 J
The negative sign indicates that the work that force and displacement have opposite directions
c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job
We have that the work of friction is equal to the change of mechanical energy
         = ΔEm
 = ΔEm
         = Emf - Emo
 = Emf - Emo
        -1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴
        ½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵
        ½ 50.0 vf² = 0.561
        vf = √ 0.561 25
       vf = 3.75 m / s