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

To find the times when the ball is 33.6 m above point O, we again use the kinematic equation:

$h = ut - \frac{1}{2}gt^2$

Substituting the known values:

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

This simplifies to:

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

Now we can use the quadratic formula to solve for $t$:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 4.9, b = -28, c = 33.6$.

Calculating the discriminant:

$b^2 - 4ac = (-28)^2 - 4 \times 4.9 \times 33.6 = 784 - 660.48 = 123.52.$

Plugging values into the quadratic formula:

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

This leads to:

$t = \frac{28 \pm 11.1}{9.8}$

Calculating the two possible values for $t$ gives:

1. $t_1 = \frac{39.1}{9.8} \approx 3.98 ext{ seconds}$
2. $t_2 = \frac{16.9}{9.8} \approx 1.73 ext{ seconds}$

Thus, the two times at which the ball is 33.6 m above O are approximately 1.73 seconds and 4 seconds.