That finds the magnitude of it, though I think this would include the motion due to Earth's gravity. Which isn't what the question wants. it wants how fast did he throw the ball. If anything you need to discount the velocity that the ball would gain due to gravity. Though I'm fairly confident in...
Well its g/y=k/m which is the same thing as i said, as I am taking x to be y, as there is no other info supplied. By the fact that omega is 2pif, then by the t equation you used in the first part, omega squared =m/k, from that you can get the time period.
Not convinced by this, but try it neverthe less. At equilibrium F=kx =mg, for it to be in equilibrium, by that meaning, m/k = x/g. so putting that into the period equation. T=2pi*sqrt(m/k) is the same as T=2pi*sqrt(x/g) which is 0.827 seconds.
Why do you get that height for the height of the dog? -where does the 8.83m come from?
What I did is found the time using s=ut+1/2at^2, solved that, got two answers, one of them negative which isn't valid.
Using this equation 10.2*t*sin60 -4.9t^2 +1.96=0, and solving gives the time as 2.0025...
Well, the energy in a spring is 1/2Kx^2. This then relates to the kinetic energy, as at the top the potential energy in the spring is all converted into gravitational potential energy. Now I think, i could be wrong on this, that the amplitude above the equilibrium point is the same as the...
Silly me, forgot the phase of the thing. Since everything else in the sin =0, then sin(-pi/4) =-1/(sqrt(2) multiply this by 0.2 and you get your answer of -0.141m.
Right, since that the max speed of the ball will be when the ball hits the floor and travels horizontal, you'll need to find the time that this takes. You have an a=0, u=0 S(vertical)=0.28m.
From this you can use s=ut+1/2at2 Since U=0 the time that this takes is equal to √2s/a.
You now...
As the above post said, the displacement is 0. This is because at x=0, t= 0, so everything in the sin function becomes 0. Sin of 0 is 0. Hence the displacement =0