Physics: conservation of energy and springs

[InFiNiTeX]

Two children are playing a game in which they try to hit a small box on the floor with a marble fired from a spring-loaded gun that is mounted on a table. The target box is 2.20m horizontally from the edge of the table. Bobby compresses the spring 1.10cm, but the center of the marble falls 27.0cm short of the center of the box. How far should Rhonda compress the spring to score a direct hit? Assume that neither the spring nor the ball encounters friction in the gun.

i know i use conservation of energy and X=vt+(1/2)Axt^2 and Y=Voy(t)+(1/2)Ayt^2 and i think you solve for both... but after that i don't know what to do... am i using the right equations? the answer is 1.25cm

 

[Doc Al]

Realize that the time it takes for the marble to fall depends only on the height so is a constant. Thus the horizontal distance the marble travels is proportional to its horizontal speed. Now relate the horizontal speed to the amount of spring compression using conservation of energy: $1/2 k x^2 = 1/2 m v^2$. Use ratios.

 

[InFiNiTeX]

i do not really understand... i was told i would most probably use conservation of energy and the 2d motion equations.. is that the same as what you explained?

 

[Doc Al]

Realize that there are usually several ways--all equivalent--to solve these kinds of problem. I don't know what you mean by "2d motion equations"; I assume you mean the kinematic equations you wrote? If so, realize that only the vertical motion is accelerated; the horizontal speed remains constant.

Bottom line: Yes, use conservation of energy and kinematics of projectiles.