Suppose you throw a rock nearly straight up at a coconut in a palm tree and the rock just misses the coconut on the way up but hits the coconut on the way down. Neglecting air resistance and the slight horizontal variation in motion to account for the hit and miss of the coconut, how does the speed of the rock when it hits the coconut on the way down compare with what it would have been if it had hit the coconut on the way up?

This analysis exercise needs us to find the speed of the coconut in the two positions, when it is above, near the tip of the palm that has a height h and when it goes back down at the same point.

Let's calculate the speed when it goes up.

Let's use mechanical energy conservation

Initial (lower part)

Em₀ = K = ½ m v²

Final (High part)

[tex]Em_{f}[/tex] = K + U = ½ m vf² + m g h

We are assuming that the point where the coconut is is not the maximum height

Emo = [tex]Em_{f}[/tex]

½ m v₀² = ½ mv² + m g h

We calculate

v² = vo2 - 2 g h

v = √ (vo² -2 g h)

If we relate the initial speed to the maximum height we fear

vo² = 2 g hmax

When replacing we have left

v = √ (2g ([tex]h_{max}[/tex] - 2gh)

v= √(2g ([tex]h_{max}[/tex -h))

We can see that the lower speed than the one that came out, the higher the palm is, the lower the velocity is parel point h = vo2 / 2g the speed of the stone is zero

Case 2 when the stone goes down and hit the coconut

Highest point

Em₂ = mg [tex]h_{max}[/tex

Lowest point (ground)

Em₃ = K + U = ½ m v₃² + mg h

Em₂ = Em₃

mg [tex]h_{max}[/tex = ½ m v₃² + mg h

v₃² = 2 g ([tex]h_{max}[/tex -h)

v₃ = √ 2g ([tex]h_{max}[/tex - h)

At this point the speed of the stone increases as the stone falls, the lower the coconut the faster it hits

Let's compare the two speeds:

- To hit the same coconut going up the speed is decreasing and if it hits you going down the speed is increasing, it has the same module

- The direction is different when it goes up is up when it goes down to down.