The kinetic energy of a moving object is directly proportional to its mass and directly proportional to the square of its velocity. This means that an object with twice the mass and equal speed will have twice the kinetic energy while an object with equal mass and twice the speed will have quadruple the kinetic energy.

So, an object traveling at a speed of about 86.6% of the speed of light will have twice as much total relativistic energy as it does when it is stationary.

Kinetic energy is directly proportional to the mass of the object and to the square of its velocity: K.E. = 1/2 m v2. If the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

The root mean square velocity is the square root of the average of the square of the velocity. As such, it has units of velocity. The reason we use the rms velocity instead of the average is that for a typical gas sample the net velocity is zero since the particles are moving in all directions.

Therefore, at the highest or at maximum height, the net velocity is u= vocosθ. So, we can say that at the maximum height, the kinetic energy is minimum as the vertical velocity is zero. Minimum kinetic energy = 12m(v0cosθ)2 . So, the correct answer is “Option D”.

When the ball is at maximum height (hmax), its velocity is zero; therefore KE = 0 and PE = mghmax. If energy is conserved, the initial kinetic energy is equal to the final potential energy (KEi = PEf).

So the kinetic energy at the maximum height is . When a projectile is at the maximum height in its trajectory, the vertical component of its velocity (for that moment) is zero. So, the potential energy at the top-most point is equal to the kinetic energy contribution by the vertical component of initial velocity.

30 meters

The maximum height will occur at the vertex of the parabola. We can find that maximum height by first finding the amount of time it takes to reach that height, then plug that time into the height function. To find the value of t we know it will occur at -b/(2a). That gives us t =-46/(2*(-16)) = 1.4375 seconds.

4 seconds

The maximum height of the object is the highest vertical position along its trajectory. The maximum height of the projectile depends on the initial velocity v0, the launch angle θ, and the acceleration due to gravity. The unit of maximum height is meters (m). H = maximum height (m)

12.4 meters

t = -b 2a = -120/(-32) = 3.75 The height of the rocket is a maximum 3.75 seconds after its launch. To find the height we only need to plug into the model: s(3.75) = -16(3.752) + 120(3.75) + 80 = 305 The maximum height of the rocket is 305 feet.

mechanical energy

From calculus, the formula is (0.5)kx^2, where x^2 is the square of the initial displacement of the end of the spring. The kinetic and potential energy at any point will sum to this value. Identify the spring’s maximum kinetic energy, at the equilibrium point, as equal to the initial potential energy.

When the kinetic energy is maximum, the potential energy is zero. This occurs when the velocity is maximum and the mass is at the equilibrium position. The potential energy is maximum when the speed is zero.

The maximum kinetic energy KEe of ejected electrons (photoelectrons) is given by KEe=hf– BE, where hf is the photon energy and BE is the binding energy (or work function) of the electron to the particular material.