A ball encounters no air resistance when thrown into the air with 100 J of kinetic energy, which is transformed to gravitational potential energy at the top of its trajectory. What is its kinetic energy when it returns to its original height

The kinetic energy when it returns to its original height is 100 J

Explanation:

The ball is thrown up with a Kinetic Energy K. E. = 0.5×m×v² = 100 J

Therefore the final height is given by

u² = v² -2·g·s

Where:

u = final velocity = 0

v = initial velocity

s = final height

Therefore v² = 2·g·s = 19.62·s

P.E = Potential Energy = m·g·s

Since v² = 2·g·s

Substituting the value of v² in the kinetic energy formula, we obtain

K. E. = 0.5×m×2·g·s = m·g·s = P.E. = 100 J

When the ball returns to the original height, we have

v² = u² + 2·g·s

Since u = 0 = initial velocity in this case we have

v² = 2·g·s and the Kinetic energy = 0.5·m·v²

Since m and s are the same then 0.5·m·v² = 100 J.

100 J

Explanation:

If there is no air resistance, then the kinetic energy is conserved. That at first the ball has 100 J of kinetic energy, when it reaches the top of its trajectory the ball has 100 J of gravitational potential energy (the kinetic energy is zero and all of them at the initial point has transformed into gravitational potential energy) , and finally when it returns to its original height it has again 100 J of kinetic energy (inverse effect than before).

if you already have the answer to (a), then just work out how many periods there are in 1 minute, and that's the answer to (c).

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s