Question

A spring-loaded gun, fired vertically, shoots a marble 7.2m straight up in the air. What is the marble's range if it is fired horizontally from 1.8m above the ground?

Answer 1

The first part means that the gun can give a amount mgh of enery to the marble. That must be the same kinetic energy it can give horizontally. \[E _{v}=E _{h}=mgh _{1}=\frac{ 1 }{ 2 }m v ^{2} \rightarrow v=\pm \sqrt{2gh _{1}}\]Now, the range is v times the time flought. That time is just the time needed for gravity to get the marble to the floor:\[x _{f}=x _{i}+ v _{i}t+\frac{ 1 }{ 2 }at ^{2} \rightarrow 0=h _{2}+0\frac{ m }{ s ^{2} }t-\frac{ 1 }{ 2 }g t ^{2} \rightarrow t=\pm \sqrt{\frac{ 2h _{2} }{ g }}\]Finally we just multiply:\[vt=\sqrt{2gh _{1}}\sqrt{\frac{ 2h _{2} }{ g }}=2\sqrt{h _{1} h _{2}}=2\sqrt{7.2m \times1.8m}\]