A ball is thrown vertically upwards and returns to its original position 2.4 s later - AQA - A-Level Physics - Question 23 - 2022 - Paper 1

Using the kinematic equation for an object in free fall:

$h = v_i t + \frac{1}{2} a t^2$

In this case, the initial velocity ($v_i$) can be derived from the equation under constant acceleration, where the final velocity at the peak is zero:

$v_f = v_i + a t$

Setting $v_f = 0$, we rearrange the equation which gives us: $v_i = -a t$ Substituting $a = -9.81 ext{ m/s}^2$ (acceleration due to gravity) and $t = 1.2 ext{ s}$ we find: $v_i = 9.81 \times 1.2 = 11.772 \text{ m/s}$

Then, substituting this back into the first equation: $h = (11.772 \text{ m/s})(1.2 \text{ s}) + \frac{1}{2}(-9.81 \text{ m/s}^2)(1.2 \text{ s})^2$ Calculating we get:

$h = 14.15 ext{ m}$. This is the maximum height reached.

The total distance travelled by the ball is the sum of the ascent and descent:

$\text{Total distance} = h + h = 14.15 ext{ m} + 14.15 ext{ m} = 28.3 ext{ m}$

Rounding to the nearest available option from the choices, the closest option is:

Answer D: 28 m.