At time $t = 0$ a ball is projected vertically upwards from a point $O$ and rises to a maximum height of 40 m above $O$ - Edexcel - A-Level Maths Mechanics - Question 1 - 2011 - Paper 1

We will use the same kinematic equation:

$h = ut + \frac{1}{2}(-g)t^2$

where:

• $h = 33.6 ext{ m}$
• $u = 28 ext{ m s}^{-1}$
• $g = 9.8 ext{ m s}^{-2}$

Substituting the known values into the equation gives:

$33.6 = 28t - \frac{1}{2} \times 9.8t^2$ $33.6 = 28t - 4.9t^2$

Rearranging gives:

$4.9t^2 - 28t + 33.6 = 0$

Using the quadratic formula, $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 4.9$, $b = -28$, and $c = 33.6$:

Calculating the discriminant: $b^2 - 4ac = (-28)^2 - 4 \times 4.9 \times 33.6$ $= 784 - 660.48$ $= 123.52$

Now substituting back into the quadratic formula:

$t = \frac{28 \pm \sqrt{123.52}}{2 \times 4.9}$ $= \frac{28 \pm 11.1}{9.8}$

This results in two possible solutions:

1. $t = \frac{39.1}{9.8} \approx 4 ext{ s}$
2. $t = \frac{16.9}{9.8} \approx 1.73 ext{ s}$

Thus, the times when the ball is 33.6 m above O are approximately 4 s and 1.73 s.