Question

The speed of a rocket just after being launched is 12 m / s.

Answer 1

This question involves the concepts of the equations of motion, kinetic energy, and potential energy.

a. The kinetic energy of the rocket at launch is "3.6 J".

b. maximum gravitational potential energy of the rocket is "3.6 J".

a. KINETIC ENERGY AT LAUNCH

The kinetic energy of the rocket at launch is given by the following formula:

$K.E=\frac{1}{2} m v_i^2$

where,

• K.E = initial kinetic energy = ?
• m = mass of rocket = 0.05 kg
• $v_i$ = initial speed = 12 m/s

Therefore,

$K.E=\frac{1}{2}(0.05\ kg)(12\ m/s)^2$

K.E = 3.6 J

b. MAXIMUM GRAVITATIONAL POTENTIAL ENERGY

First, we will use the third equation of motion to find the maximum height reached by rocket:

$2gh=v_f^2-v_i^2$

where,

• g = -9.81 m/s²
• h = maximum height = ?
• vf = final speed  = 0 m/s

Therefore,

2(-9.81 m/s²)h = (0 m/s)² - (12 m/s)²

h = 7.34 m

Hence, the maximum gravitational potential energy will be:

P.E = mgh

P.E = (0.05 kg)(9.81 m/s²)(7.34 m)

P.E = 3.6 J

Learn more about the equations of motion here:

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