A particle is projected with an initial velocity of $26 \, i + 40 \, j \, m \, s^{-1}$ from a point C on a horizontal plane - Leaving Cert Applied Maths - Question 3 - 2021

The horizontal range can be calculated by using the horizontal component of the velocity and the time of flight. The horizontal component is:

$u_x = 26 \, m \, s^{-1}$.

The range $R$ is given by:

$R = u_x \cdot t$

Thus, substituting the values:

$R = 26 \, m \, s^{-1} \cdot 8 \, s = 208 \, m$.

Using the vertical motion equation again, we can find the time at which the particle is 60 m above the horizontal. We solve:

$-60 = 40t - 5t^2$

This simplifies to the same equation solved in part (i):

$5t^2 - 40t - 60 = 0$

As previously derived, we found:

t^2 - 8t + 12 = 0
This factors to:

$(t - 6)(t - 2) = 0$ Thus, $t = 2 \, s$ and $t = 6 \, s$.

The distance between points P and Q is simply the horizontal distance traveled during the times calculated. As both points are at the same vertical height and the horizontal motion is uniform, we can use:

$|PQ| = u_x \cdot (t_2 - t_1) = 26 \, m \, s^{-1} \cdot (6 - 2) \, s = 26 \, m \, s^{-1} \cdot 4 \, s = 104 \, m$.

For the second part, we analyze the vertical motion from the cliff height. The vertical displacement $y$ is given by:

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

Here:

• $y = -98 \, m$ (the drop distance from the height of the cliff)
• $u = 35 \sin(\alpha) \, m \, s^{-1}$
• $a = -9.81 \, m \, s^{-2}$.

So the equation becomes:

$-98 = 35 \sin(\alpha) t - \frac{1}{2}(9.81)t^2$.

With $\sin(\alpha) = \frac{3}{\sqrt{13}} \Rightarrow 35\sin(\alpha) \approx 27.14$, solving the quadratic equation gives:

$t = 7 \, s$.

Finally, to find the horizontal distance $x$ when the particle lands:

$x = u_x \cdot t$

With $u_x = 35 \cos(\alpha)$, where $\cos(\alpha) = \frac{2}{\sqrt{13}}$: