During a soccer practice session, the football is positioned at the...

Given, initial speed $u = 64 \text{ ft/s}$

Angle of projection \theta = 45^{}^{{}^{\circ}}

Horizontal distance to goalposts $x = 40 \text{ yards} = 40 \times 3 = 120 \text{ ft}$

Acceleration due to gravity $g = 32 \text{ ft/s}^2$

The equation of trajectory is given by:

$y = x \tan \theta - \dfrac{g x^2}{2 u^2 \cos^2 \theta}$

Substituting the values:

y = 120 \tan 45^{}^{{}^{\circ}} - \dfrac{32 \times (120)^2}{2 \times (64)^2 \cos^2 45^{}^{{}^{\circ}}}

$y = 120 (1) - \dfrac{32 \times 14400}{2 \times 4096 \times \dfrac{1}{2}}$

$y = 120 - \dfrac{14400}{128}$

$y = 120 - 112.5 = 7.5 \text{ ft}$

Since the height of the ball is $7.5 \text{ ft}$, which is less than the goalpost height of $10 \text{ ft}$, the ball reaches the goalposts at a height of $7.5 \text{ ft}$.