A 0.145 kg rock is thrown with a speed of 32.0 m/s at an angle of 400. what is its kinetic energy at the top of its trajectory?



Answer :

The kinetic energy at the top of the rock trajectory is 43.57 Joules

The kinetic energy formula and the procedure we will use is:

K.E. = 1/2 m (v*cosθ)²

Where:

  • v = speed
  • m = mass
  • θ = angle
  • Joules = Kg m²*s²

Information of the problem:

  • v = 32 m/s
  • m = 0.145 Kg
  • θ = 40°

Applying kinetic energy formula we get:

K.E. = 1/2 m (v*cosθ)²

K.E. = 1/2 * 0.145 kg (32 m/s*cos40°)²

K.E. = 1/2 * 0.145 kg *(24.513 m/s)²

K.E. = 1/2 * 0.145 kg *600.91 m²/s²

K.E. =  87.132 kg m²/s² / 2

K.E. =  43.57 Joules

What is kinetic energy?

It is the energy possessed by a body due to its relative motion. It is usually expressed in Joules (J).

Learn more about kinetic energy at brainly.com/question/114210

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