A ball is thrown from a point A at a target T, which is on horizontal ground - Leaving Cert Applied Maths - Question 3 - 2016

To find the horizontal distance from point A to point B, we first need the horizontal component of the initial velocity:

$v_{0x} = 25 \cos(30°)$

Using the time calculated from part (i):

$x = v_{0x} t$

where $t$ is the time taken calculated in part (i). Thus,

$TB = \left( 25 \cos(30°) \right) t.$

This needs to be calculated using the total time.

The speed of the ball at point C can be found by considering the components of the velocity.

1. The vertical component of the velocity at point C is: $v_y = v_{0y} + gt$
2. The horizontal component remains the same: $v_x = v_{0x}$
3. The total speed at point C can be calculated using: $|v| = \sqrt{v_x^2 + v_y^2}$ Plugging in the values calculated from the previous parts to find the exact speed.

Using the formula for the maximum range:

$R = \frac{ku^2}{g}$

We also know that the angle of projection $\theta$ and the inclined plane will affect this range. To find $k$, we need to derive the formula based on the motion equations and substitute the known values to isolate $k$ for final calculation.