In this question use $g = 9.81 \, \text{ms}^{-2}$ A particle is projected with an initial speed $u$, at an angle of $35^{\circ}$ above the horizontal - AQA - A-Level Maths Mechanics - Question 16 - 2018 - Paper 2

For the total time of flight, we use:

$s = ut + \frac{1}{2}at^2$

For vertical motion, we take:

• Displacement $s = -10 \, \text{m}$ (downward),
• Initial velocity $u = 25.7 \sin(35^{\circ})$,
• Acceleration $a = -9.81 \, \text{ms}^{-2}$
• Total time $t$.

We know:

1. First, calculate $u \sin(35^{\circ})$:

$u \sin(35^{\circ}) = 25.7 \times 0.5736 \approx 14.71 \, \text{ms}^{-1}$

2. Substituting values into the equation:

$-10 = 14.71t - \frac{1}{2}(9.81)t^2$

3. Rearranging gives:

$4.905t^2 - 14.71t - 10 = 0$

4. Using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where:

• $a = 4.905$,
• $b = -14.71$,
• $c = -10$.

Calculating:

$t = \frac{14.71 \pm \sqrt{(-14.71)^2 - 4 \times 4.905 \times (-10)}}{2 \times 4.905}$

5. Solving this yields two values for $t$. However, we only consider the positive time, approximately:

$t \approx 3.57 \, \text{s}$

Thus, the total time the particle is in flight is approximately:

$t \approx 3.57 \, \text{s}$.