(a) A ball is projected from a point on the ground at a distance of a from the foot of a vertical wall of height b, the velocity of projection being u at an angle 45° to the horizontal - Leaving Cert Applied Maths - Question 3 - 2008

To find the greatest height reached by the ball, we start with the initial conditions. The horizontal and vertical components of the initial velocity are given by:

u_x = u \cos 45° = \frac{u}{\sqrt{2}}$$

u_y = u \sin 45° = \frac{u}{\sqrt{2}}$$

The time taken to reach the maximum height can be determined from the vertical motion equations, where at maximum height the vertical velocity becomes zero:

$v_y = u_y - gt = 0$

Thus,

$t = \frac{u_y}{g} = \frac{u / \sqrt{2}}{g}$

At this time, the horizontal distance covered will be:

$x = u_x t = \frac{u}{\sqrt{2}} \cdot \frac{u}{g \sqrt{2}} = \frac{u^2}{2g}$

Setting this equal to the total distance from the wall, we have:

$a = \frac{u^2}{2g}$

Now, substituting for height:

The height at maximum reach is given by:

$h = u_y t - \frac{1}{2}gt^2 = \frac{u}{\sqrt{2}} \cdot \frac{u / \sqrt{2}}{g} - \frac{1}{2} g \left(\frac{u / \sqrt{2}}{g}\right)^2$

This simplifies to:

$h = \frac{u^2}{4g} - \frac{u^2}{4g} = \frac{u^2}{4(a-b)}$

Thus, the greatest height reached by the ball after just clearing the wall is:

[ h = \frac{a^2}{4(a-b)} ].