The escape speed from a very small asteroid is only 24 m/s. If you throw a rock away from the asteroid at a speed of 30 m/s, what will be its final speed? Entry field with incorrect answer 25.50 m/s

Given Information:  

Initial speed of rock = vi = 30 m/s  

escape speed of the asteroid = ve = 24 m/s  

Required Information:  

final speed of rock = vf = ?

vf = 18 m/s

Explanation:  

As we know from the conservation of energy

KEf + Uf = KEi + Ui

Where KE is the kinetic energy and U is the potential energy

0 + 0 = ½mve² - GMm/R

When escape speed is used, KEf is zero due to vf being zero. Uf is zero because the object is very far away from mass M, therefore, the equation becomes

GMm/R = ½mve²

m cancels out

GM/R = ½ve²

GM/R = ½(24)²

GM/R = 288

KEf + Uf = KEi + Ui

½mvi² + 0 =  ½vf² - GMm/R

m cancels out

½vi² =  ½vf² - GM/R

Substitute the values

½(30)² =  ½vf² - (288)

½vf² = 450 - 288

vf² = 2(162)

vf = √324

vf = 18 m/s

Therefore, the final speed of the rock is 18 m/s

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what is the actual question though?