QUESTION IMAGE
Question
a quarterback claims that he can throw the football a horizontal distance of 178 m. furthermore, he claims that he can do this by launching the ball at the relatively low angle of 25.7° 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. current attempt in progress number units attempts: 0 of 3 used submit answer using multiple attempts will impact your score. 20% score reduction after attempt 2 save for later
Step1: Analyze horizontal - motion
The horizontal - motion of a projectile is a uniform - motion with the formula $x = v_{0x}t$. The initial horizontal velocity $v_{0x}=v_0\cos\theta$, where $x = 178$ m and $\theta = 25.7^{\circ}$. So $x = v_0\cos\theta\times t$, and $t=\frac{x}{v_0\cos\theta}$.
Step2: Analyze vertical - motion
The vertical - motion of a projectile is a uniformly - accelerated motion with $y - y_0=v_{0y}t-\frac{1}{2}gt^{2}$, and since $y - y_0 = 0$ (launched and caught at the same vertical level), $v_{0y}=v_0\sin\theta$. So $0 = v_0\sin\theta\times t-\frac{1}{2}gt^{2}$. Factoring out $t$ (we ignore $t = 0$ which is the initial time), we get $t=\frac{2v_0\sin\theta}{g}$.
Step3: Equate the two expressions for time
Set $\frac{x}{v_0\cos\theta}=\frac{2v_0\sin\theta}{g}$. Cross - multiply to get $v_0^{2}=\frac{xg}{2\sin\theta\cos\theta}$. Using the double - angle formula $\sin2\theta = 2\sin\theta\cos\theta$, we have $v_0=\sqrt{\frac{xg}{\sin2\theta}}$.
Step4: Substitute the values
Given $x = 178$ m, $g = 9.8$ m/s², and $\theta = 25.7^{\circ}$, then $2\theta=51.4^{\circ}$, and $\sin(2\theta)=\sin(51.4^{\circ})\approx0.782$. Substitute into the formula: $v_0=\sqrt{\frac{178\times9.8}{0.782}}=\sqrt{\frac{1744.4}{0.782}}\approx\sqrt{2230.7}\approx47.2$ m/s.
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$47.2$ m/s