Question

What is the speed of a 4.50-kg cannon ball if, at a height of 275 m above the ground, its mechanical energy relative to the ground is 6.27 104 J

Answer 1

The speed of the cannonball is approximately 166.9 m/s.

Step-by-step explanation:

The speed of a cannonball can be determined by considering its mechanical energy relative to the ground. In this case, the mechanical energy is given as 6.27 x 10^4 J. This energy can be equated to the kinetic energy of the cannonball, which is given by the equation:

KE = 0.5 * m * v^2

Where KE is the kinetic energy, m is the mass of the cannonball, and v is its velocity. To solve for v, we rearrange the equation:

v = sqrt(2 * KE / m)

Substituting the given values:

v = sqrt(2 * 6.27 x 10^4 J / 4.50 kg)

v = sqrt(2 * 13933.33 m^2/s^2)

v = sqrt(27866.67 m^2/s^2)

v = 166.9 m/s

Therefore, the speed of the cannonball is approximately 166.9 m/s.