Question #10940

An object of mass 0.5 kg is thrown vertically upward with a speed of 4 m/s.
i) Calculate the kinetic energy of the object.
ii) What is the maximum potential energy that the object will possess?
iii) Hens,calculate the maximum height reached by the object?

Expert's answer

An object of mass 0.5 kg is thrown vertically upward with a speed of 4 m/s.

i) Calculate the kinetic energy of the object.

The initial kinetic energy of an object is

Ek = mV²/2 = 0.5·4²/2 = 4 J.

ii) What is the maximum potential energy that the object will possess?

The sum of kinetic and potential energy is constant. The kinetic energy transforms to potential energy, and at the highest point of the object's trajectory the kinetic energy is equal to zero. The maximum potential energy will be reached at that point, so

max(Ep) = 4 J.

iii) Hence, calculate the maximum height reached by the object?

h(t) = Vt - gt²/2

h'(t) = V - gt

h'(t) = 0 <==> V = gT ==> T = V/g = 4/9.8 ≈ 0.4082 s

H = max(h(T)) ≈ h(0.4082) = 4·0.4082 - 9.8·0.4082²/2 ≈ 0.8163 m.

i) Calculate the kinetic energy of the object.

The initial kinetic energy of an object is

Ek = mV²/2 = 0.5·4²/2 = 4 J.

ii) What is the maximum potential energy that the object will possess?

The sum of kinetic and potential energy is constant. The kinetic energy transforms to potential energy, and at the highest point of the object's trajectory the kinetic energy is equal to zero. The maximum potential energy will be reached at that point, so

max(Ep) = 4 J.

iii) Hence, calculate the maximum height reached by the object?

h(t) = Vt - gt²/2

h'(t) = V - gt

h'(t) = 0 <==> V = gT ==> T = V/g = 4/9.8 ≈ 0.4082 s

H = max(h(T)) ≈ h(0.4082) = 4·0.4082 - 9.8·0.4082²/2 ≈ 0.8163 m.

## Comments

## Leave a comment