If a footballer kicks a ball straight down the pitch at 6 ms-1 at an angle θ of 30° above the horizontal, what is the maximum height reached by the ball?

First we must remember our equations of motion, (SUVAT equations).S = Displacement U = Initial velocity V = Final velocity A = Acceleration T = TimeV = U + ATS = UT + (1/2)AT² V² = U² + 2ASS = (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^-1Horizontal intial velocity = 6ms^-1Cos(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