1. The problem statement, all variables and given/known data In my physics 12 class, we were given a spring and had to launch it at an angle into a target (the object launched is the spring itself). The target was on the floor level but the spring was launched from above the floor level. I need to find out how far to pull the spring back so it will travel to the target. I also have to find out how to calculate the maximum height of the spring. Mass of spring = 0.04012kg k = 26 N/m (it was measured before ) Height of launch from the ground: 0.48 m The horizontal distance from the target to the spring: 4 m 2. Relevant equations It is clearly related to the elastic potential of the spring and kinetic energy as well. Es = Ek 1/2mv^2 = 1/2kx^2 v^2 = x^2(k/m) v = x√(k/m) Beacuase we dont have the velocity, we have to work with projectile motion as well. Δd = (v)(cosΘ)(t) ==> re-arranging => t = (Δd) / (v cosΘ) five kinematics equations might be needed but im not sure. 3. The attempt at a solution Because we dont have velocity, we sure have to work with kinematics as well as energy. I dont know how to relate them together to get the x (distance needed to be pulled). There are some websites that explain a bit but jumped right into the equation. such as: http://04adams.tripod.com/physics/labs/term2/spring_lab/ http://www.geocities.ws/juliars123/springs.htm They both got the equation: x = √((dmg)/(2k*sinΘ*cosΘ)). i used this equation and it worked. but i dont get how they got it. in specific, both of them use this equation : d = (V*cosΘ*2V*sinΘ)/g to get the final equation, which they dont explain how they got it. Again, i dont know for sure if these two websites are right. I just would like someone to help me get the right equation or explain the equation achieved by the above websites.