Question

A projectile of mass 2 kg is fired at an angle to the ground such that its flight is parabolic. At a certain point in its flight, its potential energy is 150 J and its kinetic energy is 250 J. What is the speed of the projectile when it hits the ground? You may assume that the potential energy at ground level is zero and there is no air resistance?

Answer #1

Mass = 2kg

At a certain point,

Potential energy PE_{1} =150J

Kinetic Energy KE_{1} = 250J

Total Energy T = Potential Energy + Kinetic energy

T= 150+250 = 400J

According to conservation of energy, total energy will remain constant.

At the point where the mass hits the ground,

Potential energy is zero

Kinetic Energy = K

Total energy T = Potential Energy + Kinetic Energy

T= 0+K

Since total energy is 400J,

K=400J

This is the kinetic energy of the mass when it hits the ground. Let its speed at that point be v, therefore:-

mv^{2}/2 =k

2v^{2}/2 = 400J

v^{2} = 400

v = 20 m/s

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