Suppose Captain Marvel leaps from the top of a 20 m tall building. If his initial speed is 12 m/s, then what is his speed just before he lands on the ground?
To find Captain Marvel's speed just before he lands on the ground, we can use the principle of conservation of energy.
The initial potential energy of Captain Marvel at the top of the building will be equal to the kinetic energy just before he lands on the ground.
The potential energy (PE) of an object at a certain height is given by the formula:
PE = m * g * h
where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
In this case, we can assume Captain Marvel's mass is negligible since he is a fictional character made of superpowers, and we only need to consider the change in potential energy.
The initial potential energy (PE_initial) is:
PE_initial = m * g * h
PE_initial = 0 * 9.8 * 20
PE_initial = 0 J
Since there is no initial potential energy, all of the energy is converted into kinetic energy just before Captain Marvel lands.
The kinetic energy (KE) of an object is given by the formula:
KE = (1/2) * m * v^2
where v is the speed of the object.
Setting the initial potential energy equal to the kinetic energy, we have:
0 = (1/2) * m * v^2
Since the mass cancels out, we can simplify the equation to:
0 = (1/2) * v^2
Multiplying both sides by 2, we have:
0 = v^2
Taking the square root of both sides, we get:
v = 0 m/s
Therefore, Captain Marvel's speed just before he lands on the ground is 0 m/s.