A particle is projected with initial velocity $40 \, extbf{i} + 50 \, extbf{j} \, \text{m/s}$ from point p on a horizontal plane - Leaving Cert Applied Maths - Question 3 - 2009

The speed of the particle can be calculated as: $\text{speed} = \sqrt{(40)^2 + (30)^2}$

Calculating the speed: $\text{speed} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, \text{m/s}$

To find the direction, we compute the angle using: $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{30}{40}$

This gives: $\theta = \tan^{-1} \left(\frac{3}{4}\right) \approx 36.87^\circ$

Using the horizontal motion equation, we know that: $\text{horizontal distance} = \text{initial horizontal velocity} \times \text{time}$;

after 2 seconds:

$360 = 40(2)$

Then we calculate k using the vertical motion: $s_y = ut + \frac{1}{2} a t^2$ Given that the vertical displacement is: $k = 50(2) - 5(2^2)$

This simplifies to: $k = 100 - 20 = 80 \text{, so } k = 45 \text{.}$

For the projectile fired horizontally, the horizontal distance traveled is given by: $s_x = ut$

Where:

• The vertical fall is: $s_y = -45$,
• Using the vertical motion equation for time: $-45 = 0t + \frac{1}{2}(-g)t^2$ This simplifies to: $45 = \frac{1}{2}(10)t^2$
• Solving gives: $t^2 = 9 \Rightarrow t = 3 \, \text{s}.$

The horizontal distance can now be expressed as: $s_x = x(3) = \frac{30}{\sqrt{3}}$

Thus, $x = \frac{30}{\sqrt{3} \cdot 3} \Rightarrow x = 10 \text{ m/s}.$