1. The problem statement, all variables and given/known data A ball of mass m is launched at an angle theta from a spring gun of length D. The lower end of the gun, where the spring is attatched, is at ground level. The spring has spring constant k and an unstretched length D. Before launch, the spring is compressed to a length d. Determine the maximum height h above the ground the ball reaches. 2. Relevant equations (k(x^2)/2)=mgh ?? x = mgcos(theta) 3. The attempt at a solution Plugging in "x" I get: h=(k(mgcos(theta))^2)/2mg Is this correct? If not please explain in detail and provide the steps and solution. TIA.
Here x is a velocity. You can use conservation of energy to find the additional height reached after the ball leaves the gun, if x is the vertical component of the speed at that point. Note that h will not be the height above the ground. You can't set a velocity (x) equal to a force. No. Note that you haven't even used the spring constant or the amount the spring was compressed. Try to solve it in two steps: Find the speed of the ball as it leaves the gun. Then find the height that it reaches after it becomes a projectile.
Well then I guess I have no idea what i'm doing then...A little more help please K_i + U_i + S_i = K_f + U_f + S_f K = .5mv^2 U = mgh S = .5kx^2 What is the muzzle velocity? How to find the height that it reaches?
To find the muzzle velocity, use conservation of energy as you outlined above. Hint: What's the change in height of the ball as it travels through the gun? Once the ball leaves the gun it's a projectile. Solve that part like any other projectile motion problem.
Could someone please solve the problem for me.....BTW this is NOT a homework problem....It is a final test review problem....I learn differently than most.....By looking at the steps involved in solving the problem along with the answer I can use deductive reasoning as to why those methods were used....TIA