In this question use $g = 9.8 \, \text{ms}^{-2}$ A ball is projected from a point on horizontal ground with an initial velocity of $7 \, \text{ms}^{-1}$ at an angle $\theta$ above the horizontal - AQA - A-Level Maths Pure - Question 13 - 2022 - Paper 2

To find the maximum vertical height $h$, we can use the kinematic equation:

$v^2 = u^2 + 2as$

where:

• $v$ is the final vertical velocity at the maximum height (which is 0).
• $u$ is the initial vertical component of velocity.
• $a$ is the acceleration due to gravity, which is negative.
• $s$ is the vertical displacement, which is $h$ in this case.

First, we calculate the initial vertical component of the velocity:

$u = 7 \sin \theta$

Substituting everything into the kinematic equation gives:

$0 = (7 \sin \theta)^2 + 2(-9.8)h$

So, we rearrange it as follows:

$0 = 49 \sin^2 \theta - 19.6h$

Therefore,

$h = \frac{49 \sin^2 \theta}{19.6} = 2.5 \sin^2 \theta.$