Using the equation;
TE = 1/2mv^2(1+2); where k = 2/5 for a solid sphere; V is the velocity, and m is the mass.
Total energy = 0.5 × 21 × 8² (7/5)
= 940.8 J
The rotational kinetic energy of the sphere is 940.8 J
To solve this problem we will apply the concept related to the moment, which describes the change in speed in proportion to the body mass. The momentum can be described under the general equation

Where,
m = mass
v = Velocity
The change in momentum is the difference between the final moment and the initial moment

Where,
Final momentum
Initial momentum
After catching the ball, the ball comes to rest position and speed of the ball is zero.

The change in momentum on catching the ball is


The momentum during the catching of the ball is

Now during deflection of the ball, we know that there is an initial momentum and a negative final momentum because it moves with the same speed but in the opposite direction, that is

The momentum is negative since during deflection the ball moves with same speed in opposite direction


Therefore, the correct ansswer is B: catch the ball in order to minimize your speed on the skateboard
Explanation:
It is given that, the water from a fire hose follows a path described by equation :
........(1)
The x component of constant velocity, 
We need to find the resultant velocity at the point (2,3).
Let
and 
Differentiating equation (1) wrt t as,



When x = 2 and 
So,


Resultant velocity, 

v = 6.4 m/s
So, the resultant velocity at point (2,3) is 6.4 m/s. Hence, this is the required solution.