Standing at the edge of a cliff 240 m tall, you throw two balls into the air – one directly upwards at 5 m/s and another directly downwards at -5 m/s. Use conservation of energy to show that they will have the same final speed at the bottom.

Question

Standing at the edge of a cliff 240 m tall, you throw two balls into the air – one directly upwards at 5 m/s and another directly downwards at -5 m/s.  Use conservation of energy to show that they will have the same final speed at the bottom.

Elena · Answer

The 1st ball:
PE+KE₀=KE
mgH+mv₀²/2 =mv1²/2
v1=sqrt(2gH+ v₀²)
The 2nd ball:
upwards motion
KE1=PE1
mv₀²/2 =mgh
h= v₀²/2g.
downwards motion:
PE2=KE2
mg(H+h) = mv2²/2
v2=sqrt{ 2g(H+h)}=
= sqrt{ 2g(H+ v₀²/2g )}=
= sqrt {2gH + v₀²}
=> v2=v1

Scoop · Answer

Could you break it down a little further?