A ball is launched from ground level at 5m/s at an angle of 30 degrees above the horizontal. What is its height above ground level at the highest point in its trajectory?

STEP 1: Draw a diagram!

Remember to include all the useful information you are given in the question i.e. the initial velocity of the ball (5m/s at 30 degrees above the horizontal). This can be shown as a vector quantity pointing at 30 degrees above a horizontal reference line, with horizontal and vertical components shown in dotted lines.

Add your sign convention to the diagram. This is shown by drawing an arrow pointing to each of the positive horizontal and vertical directions. It doesn’t matter what sign convention you choose, just stick to it.

It may also be useful to draw a rough trajectory for the ball (an upside-down parabola) to visualise what is going on.

STEP 2: Set up your SUVAT equations

Write down the information you have in terms of the SUVAT equation quantities. Remember that the horizontal and vertical components are independent, and think about which set(s) of components you need to answer the question.

Since the question asks for the height of the ball, you only need to think about the vertical components.

So, considering only the vertical components, at the highest point in the ball's trajectory:

NOTE: I used a sign convention such that the positive vertical direction is upwards.

s = displacement = H (This is the quantity you want to find, so leave it as an unknown. I've called it H (for height) here.)

u = initial velocity = 5*sin(30) = 2.5m/s (Remember to only use the vertical component of the initial velocity! This was found from the vector diagram earlier, using trigonometry.)

v = velocity of ball at time of interest = 0m/s (This is key to getting the correct answer. At the instant that the ball reaches its highest point, it cannot be moving any higher, but has not yet started moving downwards, so its vertical velocity must be momentarily zero. Therefore V=0m/s.)

a = acceleration = -g = -9.81m/s^2 (The acceleration everywhere on the surface of Earth is a constant 9.81m/s^2, given the symbol “g” for convenience. Exam Tip: Take “g” to be 9.81m/s^2, unless the question says otherwise.)

t = time elapsed = ???? (You don’t need to know the time elapsed to answer the question so ignore T!)

Substitute these values into the appropriate SUVAT equation ( v^2 = u^2 + 2as ) to get: 0 = 2.5^2 + 2*(-g)*H

STEP 3: Rearrange and solve!

You’ve done the hard work! Now just solve the equation:

0 = 2.5^2 – 2gH è 2gH = 2.5^2 è H = (2.5^2)/(2g) = 6.25/(2*9.81) = 0.32m (2 decimal places)