Awater baloon at m=300g is launched horizontaly from a platform 2 m above the ground with a slingshot. the sling shot (k=60n/m) is streched 40cm before launch. how far from the platform will the balloon strike the ground?

I'm assuming no air friction here, first use conservation of energy to find velocity, so initially all the energy is from the spring, (1/2)kx^2, then it is converted to kinetic energy, (1/2)mv^2. so .5kx^2=.5mv^2. the .5s cancel, kx^2=mv^2. solve for v, v=sqrt(kx^2/m) plug in your values for the variables and find v. then use vf^2=vi^2+2ax in the u direction and solve for vf, vf=sqrt(vi^2+2ax). use that velocity in vf=vi+at and solve for t. vi=0 and a=g=9.81, so vf/a=t. plug in the values from the previous equation and the acceleration due to gravity to find the time it takes to hit the ground vertically. Now use v=x/t and solve for x. x=vt, plug in the velocity of the ball horizontally, which we found with the conservation of energy part, and y which we found with the previous equation, and you get your x, the distance it will take for the ball to hit the ground.

You would need a laid out graph for this but could you provide a picture of just a graph? i can give stepssolve for y.subtract 4xdivide by -1 (opposites)graph slope intercept formcheck each point to see if it’s on the graph