Explanation:
Let's calculate the components of the football's velocity:
[tex]v_{0x} = (23.9\:\text{m/s})\cos{51.5°} = 14.9\:\text{m/s}[/tex]
[tex]v_{0y} = (23.9\:\text{m/s})\sin{51.5°} = 18.7\:\text{m/s}[/tex]
a) The time it takes for the football to travel 36.0 m horizontally is
[tex]t = \dfrac{x}{v_{0x}} = \dfrac{36.0\:\text{m}}{14.9\:\text{m/s}} = 2.4\:\text{s}[/tex]
During this time, the y-displacement of the football is
[tex]y = v_{0x}t - \frac{1}{2}gt^2[/tex]
[tex]\:\:\:\:= (18.7\:\text{m/s})(2.4\:\text{s}) - \frac{1}{2}(9.8\:\text{m/s}^2)(2.4\:\text{s})^2[/tex]
[tex]\:\:\:\:= 16.7\:\text{m}[/tex]
This means that the football cleared the crossbar by 16.7 m - 3.05 m = 13.7 m
b) To determine whether the football was rising or falling while clearing the crossbar, let's look at the y-component of its velocity after 2.4 s:
[tex]v_y = v_{0y} - gt = 18.7\:\text{m/s} - (9.8\:\text{m/s}^2)(2.4\:\text{s})[/tex]
[tex]\:\:\:\:\:\:= -4.82\:\text{s}[/tex]
Since its sign is negative, this means that the football was already on its way down.
