Detailed PLZ! A man is standing on a building 30 m high holding a 1 kg rock. What is the potential
energy of the rock? The man drops the rock. What is the kinetic energy of the rock
the instant before it hits the ground? g=9.8m/s^2

Potential Energy = 294J, Kinetic Engergy = 48.02J

Explanation:

We have these formulas:

Potential Energy = mass * gravitational force * height (m) = 1 * 9.8 * 30 = 294(J)

Kinetic Energy = 1/2 * mass * velocity^2 = 1/2 * 1 * 9.8^2 = 48.02 (J)

As the rock falling at an acceleration of 9.8m/s^2 which means for each second, the rock increases 9.8m/s. I think we are missing time to find the instantaneous velocity, the formula is (initial displacement - final displacement)/ (initial time - final time) which will directly give the final answer for you.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

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