A particle is projected from a point P with speed u m s^-1 at an angle α to the horizontal - Leaving Cert Applied Maths - Question 3 - 2020

From the vertical motion, we have:

$y = u\sin\alpha t - \frac{1}{2}gt^2$

At ( t = 5 , s ):

$24.5 = u\sin\alpha \cdot 5 - \frac{1}{2}g(5)^2$

Using the value of g (approximately 9.81 m/s²), we can rearrange this to find u in terms of sin(α). The substitution of known values leads us to solve for u.

The initial velocity down the plane can be calculated as follows:

Given:

• Initial speed = ( 2\sqrt{2} , m , s^{-1} )
• Angle of projection = 45°

From the formula for the range down an incline: $d = \frac{2v^2 \sin 45° \cos 45°}{g}$

Substituting the initial speed will provide the required result for d.

The second hop range utilizes the coefficient of restitution. Thus, using: $kd = \left(\frac{1}{g}\right)\left(k \cdot 16\right)^2 \sin 45°\sin 45°$ By substituting the computed d and using the known coefficient, we find the value of k through rearrangement.