Question

In U.S. football, after a touchdown the team has the opportunity to earn one more point by kicking the ball over the bar between the goal posts. The bar is 3.00 m above the ground, and the ball is kicked from ground level, 11.0 m horizontally from the bar
If the ball is kicked at 44.0∘ above the horizontal, what must its initial speed be if it to just clear the bar? Express your answer in m/s.

Answer 1

Question

Motion in horizontal direction is described by the equation: $x = v_0 \cdot \cos 44{}^\circ \cdot t$, where $t$ is time of motion and $v_0$ is initial velocity. From this equation we can find time $(t)$:

$$x = v_0 \cdot \cos 44{}^\circ \cdot t \Rightarrow t = \frac{x}{v_0 \cdot \cos 44{}^\circ}.$$

Motion in vertical direction is described by the equation: $y = v_0 \cdot \sin 44{}^\circ \cdot t - \frac{g \cdot t^2}{2}$. We know time from previous equation, so we will have:

$$\begin{aligned} y = v_0 \cdot \sin 44{}^\circ \cdot \frac{x}{v_0 \cdot \cos 44{}^\circ} &= \frac{g}{2} \cdot \left(\frac{x}{v_0 \cdot \cos 44{}^\circ}\right)^2 = \\ &= x \cdot \tan 44{}^\circ - \frac{g \cdot x^2}{2 \cdot \cos^2 44{}^\circ} \cdot \left(\frac{1}{v_0}\right)^2 \Rightarrow \frac{g \cdot x^2}{2 \cdot \cos^2 44{}^\circ} \cdot \left(\frac{1}{v_0}\right)^2 = x \cdot \tan 44{}^\circ - y \Rightarrow \\ &\Rightarrow \left(\frac{1}{v_0}\right)^2 = \frac{2 \cdot \cos^2 44{}^\circ \cdot (x \cdot \tan 44{}^\circ - y)}{g \cdot x^2} \Rightarrow \frac{1}{v_0} = \sqrt{\frac{2 \cdot \cos^2 44{}^\circ \cdot (x \cdot \tan 44{}^\circ - y)}{g \cdot x^2}} \Rightarrow \\ &\Rightarrow v_0 = \sqrt{\frac{g \cdot x^2}{2 \cdot \cos^2 44{}^\circ \cdot (x \cdot \tan 44{}^\circ - y)}} \end{aligned}$$

We know that in our case $y = 3$ meters and $x = 11$ meters, $g = 9.81 \frac{m}{s^2}$. So, we will have: $v_0 = \sqrt{\frac{9.81 \cdot 11^2}{2 \cdot \cos^2 44{}^\circ \cdot (11 \cdot \tan 44{}^\circ - 3)}} = 12.27 \frac{m}{s}$.

Answer: 12.27 $\frac{m}{s}$.