# SUVAT applied to circular motion

My teacher set me this problem:

A ball is suspended from a ceiling by a string of length $$7.5\ m$$. The ball is kicked horizontally and rises to a maximum height of $$6.0\ m$$. Assuming that air resistance is negligible, show that the initial speed of the ball is $$11\ ms^{-1}$$

Now, I solved this problem by applying the SUVAT equation: $$v^2 = u^2 + 2as$$; where $$v=0$$, $$a=g$$, and $$s=6$$. Therefore $$u=11$$.

My teacher did not accept my solution and showed me how to do it using the conservation of energy: intial KE = gain in GPE; $$K_i=U_g$$. She also explained that I cannot use SUVAT because the object is in circular motion and the acceleration is not constant. Also, the acceleration $$g$$ is not in the same direction as initial speed $$u$$. I understand this.

My question is: how can I get the right answer? Both the SUVAT method and the energy conservation method produce an expression for the change in height: $$h = \frac{u^2}{2g}$$

Is this just a coincidence? One valid method and one invalid method just happen to give the same expression?

If you take your kinematic SUVAT equation (kinematics - the branch of mechanics that deals with pure motion, without reverence to the masses or forces involved in it ) $$v^2=u^2+2as$$ and multiply both sides by $$\frac 12 m$$ where $$m$$ is the mass of the mass of the ball you get $$\frac 12 m v^2= \frac 12 m u^2+mas$$ which you will see is the equation which your teacher wanted you to use.
Well let’s examine the correct method using $$\frac 12 m v^2- \frac 12 m u^2= mgs$$.
The left hand side is the change in kinetic energy and the right hand side is the work done by a force $$mg$$ when its displacement is $$s$$ along the line of action of the force which for the gravitational force $$mg$$ is in the vertical direction.
However when treating this problem correctly as a dynamics problem (dynamics - the branch of mechanics concerned with the motion of bodies under the action of forces) one has to include all forces acting on the ball and in this case as well as the force $$mg$$ acting on the ball there is the force due to the string.