A particle is projected from a point on horizontal ground with an initial speed of 82 m s⁻¹ at an angle β to the horizontal, where tan β = 4/9 - Leaving Cert Applied Maths - Question 3 - 2014

At maximum height, the vertical component of the velocity becomes zero:

$$v_y = u_y - gt.$$

Substituting values:

$$0 = 80 - 10t,\ t = 8 \text{ seconds.}$$

To calculate the maximum height ( s), we use the following kinematic equation:

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

Substituting the required values:

$$s = 80 \times 8 - \frac{1}{2} \times 10 \times (8)^2 = 320 \text{ m.}$$

The range ( R ) of the projectile can be calculated by:

$$R = v_x \times t_{total},$$

where ( t_{total} = 2t = 2 \times 8 = 16 \text{ seconds} ) and ( v_x = 18 ). Thus:

$$R = 18 \times 16 = 288 \text{ m.}$$

Using the kinematic equation:

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

we substitute ( s = 275, u = 80, a = -10 ):

$$275 = 80t - 5t^2.$$

Rearranging gives:

$$5t^2 - 80t + 275 = 0.$$

Using the quadratic formula:

$$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\ t = \frac{80 \pm \sqrt{6400 - 5500}}{10} \ = t = 5, 11.$$