PV = 400 x 0.08 = 32 J
Hope this helps
To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.
The trajectory equation from the motion kinematic equations is given by

Where,
a = acceleration
t = time
= Initial velocity
= initial position
In addition to this we know that speed, speed is the change of position in relation to time. So

x = Displacement
t = time
With the data we have we can find the time as well




With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,





Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.
Its tangential speed is constant although its velocity is changing. As the object changes direction, it results in a changing of positive and negative signs of the velocity. Although, the magnitude of the velocity (speed) is not changing.
Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km
Answer:
The moment of a given force about a given axis of rotation can be decreased by decreasing the perpendicular distance of force from the axis of rotation.