Question

a quarterback claims that he can throw the football a horizontal distance of 195 m. furthermore, he claims that he can do this by launching the ball at the relatively low angle of 32.9° above the horizontal. to evaluate this claim, determine the speed with which this quarterback must throw the ball. assume that the ball is launched and caught at the same vertical level and that air resistance can be ignored. for comparison a baseball pitcher who can accurately throw a fastball at 45 m/s (100 mph) would be considered exceptional.

Explanation:

Step1: Identify the range formula

The range formula for projectile motion when launched and caught at the same vertical - level is $R=\frac{v_{0}^{2}\sin2\theta}{g}$, where $R$ is the range, $v_{0}$ is the initial velocity, $\theta$ is the launch angle, and $g = 9.8\ m/s^{2}$ is the acceleration due to gravity.

Step2: Solve the formula for $v_{0}$

We can re - arrange the formula $R=\frac{v_{0}^{2}\sin2\theta}{g}$ for $v_{0}$. First, multiply both sides of the equation by $g$: $Rg=v_{0}^{2}\sin2\theta$. Then, divide both sides by $\sin2\theta$: $v_{0}^{2}=\frac{Rg}{\sin2\theta}$. Finally, take the square - root of both sides: $v_{0}=\sqrt{\frac{Rg}{\sin2\theta}}$.

Step3: Calculate $\sin2\theta$

Given $\theta = 32.9^{\circ}$, then $2\theta=2\times32.9^{\circ}=65.8^{\circ}$. So, $\sin2\theta=\sin(65.8^{\circ})\approx0.912$.

Step4: Substitute the values of $R$, $g$, and $\sin2\theta$

We know that $R = 195\ m$ and $g = 9.8\ m/s^{2}$. Substituting these values into the formula $v_{0}=\sqrt{\frac{Rg}{\sin2\theta}}$, we get $v_{0}=\sqrt{\frac{195\times9.8}{0.912}}$.
First, calculate the numerator: $195\times9.8 = 1911$. Then, $\frac{1911}{0.912}\approx2095.4$. And $v_{0}=\sqrt{2095.4}\approx45.8\ m/s$.

Answer:

$45.8\ m/s$