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Wile E. Coyote (Carnivorous hungribilous) is chasing the Roadrunner (Speedibus cantcatchmi) yet again. While running down the road, they come to a deep gorge, 15 m straight across and 100 m deep. The Roadrunner launches itself across the gorge at a launch angle of 15° above the horizontal, and lands with 1.5 m to spare.
(a) What was the Roadrunner's launch speed? Ignore air resistance.
(b) Wiley Coyote launches himself across the gorge with the same initial speed, but at a different launch angle. To his horror, he is short the other lip by 0.5 m. What was his launch angle? (Assume that it was lower than 15°.)

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carlos2112

Answer:

(a) [tex]vo=17.98\frac{m}{s}[/tex]

(b) [tex]\alpha =13.03779 centigrades[/tex]

Explanation:

Let´s use the maximum horizontal distance traveled by a projectile formula:

[tex]Xmax=\frac{v^{2} }{g} *sin(2\alpha )[/tex] (1)

So, we know that the maximum distance reached by the Roadrunner is:

[tex]Xmax=15+1.5[/tex]=16.5m

if we assume that:

[tex]g=9.8\frac{m}{s^{2} }[/tex]

replacing the values ​​in (1)

[tex]vo=\sqrt{\frac{16.5*9.8}{sin(30)} } =17.98\frac{m}{s}[/tex]

Using the maximum horizontal distance traveled by a projectile formula again, let´s calculate the maximum distance reached by the Coyote:

[tex]Xmax=15-0.5[/tex]=14.5m

Assuming

[tex]g=9.8\frac{m}{s^{2} }[/tex]

and replacing in (1)

[tex]sin(2\alpha )=\frac{14.5*9.8}{17.98^{2} } =0.4395564965\\\\arcsin(2\alpha)=arcsin(0.4395564965)\\\\2\alpha =26.07558729\\\\\alpha =13.03779[/tex]centigrades

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