Answer
The ball will not clear the 10-foot high goal post.
Work Step by Step
We can convert the horizontal displacement to units of meters:
$\Delta x = 45~yd\times \frac{0.9144~m}{1~yd} = 41.148~m$
We can find the time for the ball to travel this horizontal distance:
$t = \frac{d}{v_x}$
$t = \frac{d}{v_0~cos~\theta}$
$t = \frac{41.148~m}{(21~m/s)~cos~35^{\circ}}$
$t = 2.392~s$
We can find the ball's vertical displacement at this time:
$\Delta y = v_y~t+\frac{1}{2}a_yt^2$
$\Delta y = v_0~sin~\theta~t+\frac{1}{2}a_yt^2$
$\Delta y = (21~m/s)~sin~35^{\circ}~(2.392~s)+\frac{1}{2}(-9.80~m/s^2)(2.392~s)^2$
$\Delta y = 0.776~m$
When the ball has traveled a horizontal distance of 45 yards, the ball is only 0.776 meters above the ground, which is too low to make it over the 10-foot high goal post. Therefore, the ball will not clear the 10-foot high goal post.
