Projectile question involving Law of Conservation of Energy

[leighzer]

A particle of mass 0.500 kg is shot from point P, 60 m above the ground. The particle has an initial velocity Vi, with a horizontal component of 30.0 m/s. The particle rises to a maximum height of 20.0 m above P. Using the Law of Conservation of Energy, determine the vertical component of Vi.

 

[semc]

use lose in PE=gain in KE to start

 

[kyle8921]

According to the Law of Conservation of Energy, the total energy of a system remains constant. In this case, the system consists of the particle and the Earth. The initial energy of the particle at point P is purely kinetic, given by 1/2 * m * (Vi)^2, where m is the mass of the particle and Vi is the initial velocity. As the particle rises to a maximum height of 20.0 m, it gains potential energy, given by m * g * h, where g is the acceleration due to gravity and h is the height. At the maximum height, the particle has zero kinetic energy, since it has come to a momentary stop. Therefore, the total energy at this point is purely potential energy.

Equating the initial kinetic energy to the final potential energy, we can solve for the vertical component of the initial velocity, Viy. Using the given information, we get:

1/2 * m * (Vi)^2 = m * g * h
1/2 * 0.500 kg * (30.0 m/s)^2 = 0.500 kg * 9.8 m/s^2 * 20.0 m
225 m^2/s^2 = 9.8 m/s^2 * 10.0 m
225 m^2/s^2 = 98 m^2/s^2

Solving for Viy, we get:

Viy = √(225 m^2/s^2 - 98 m^2/s^2) = √(127 m^2/s^2) = 11.3 m/s

Therefore, the vertical component of the initial velocity is 11.3 m/s. This means that the total initial velocity, Vi, can be found using the Pythagorean theorem:

Vi = √((Vix)^2 + (Viy)^2) = √((30.0 m/s)^2 + (11.3 m/s)^2) = √(900 m^2/s^2 + 127 m^2/s^2) = √(1027 m^2/s^2) = 32.0 m/s

In conclusion, using the Law of Conservation of Energy, we have determined that the vertical component of the initial velocity, Viy, is 11.3 m/s and the total initial velocity, Vi, is 32.0 m/s