First we must remember our equations of motion, (SUVAT equations).S = Displacement U = Initial velocity V = Final velocity A = Acceleration T = TimeV = U + A*TS = U*T + (1/2)*A*T^{2} V^{2 }= U^{2} + 2*A*SS = (1/2)*(U+V)*T*

Then we must identify what information the question has given us.

The intial velocity (U) is 6ms^{-1} at an angle of 30° above the horizontal.

Using trigonometery we can then find the vertical and horizontal component of the velocity.(I would then use a diagram of a triangle and SOH CAH TOA to explain how to find the vertical and horizontal components)

Vertical inital velocity = 6ms^{-1} * Sin(30°) = 3ms^{-1}Horizontal intial velocity = 6ms^{-1}*Cos(30°) = 3ms*^{-1}

As we are only interested in the height the ball reaches, we will use the vertical intial velocity as our value for U.We also know the acceleration due to gravity (A) is -9.8ms^{-2} and that at its maximum height the ball will have a final velocity (V) of 0ms^{-1 }.

Using these values of V and A we can find the value of T using equation 1.

V = U + AT0 = 3 + (-9.8)*TT = 0.3 seconds*

Now we can use this value for T alongside U and V in equation 4 to find the vertical distance the ball reaches (S).

S = (1/2)(U+V)*TS = (1/2)*(3+0)(0.3)S = 0.45m

The maximum height the ball reaches is 0.45m