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3 years ago

B. Projectile on cliff (range)

Physics

1 answer:

dimulka [17.4K]3 years ago

5 0

Answer:

x = 41.28 m

Explanation:

This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.

Let's start by using trigonometry to find the initial velocity

cos 25 = v₀ₓ / v₀

sin 25 = Iv_{oy} / v₀

v₀ₓ = v₀ cos 25

v_{oy} = v₀ sin 25

v₀ₓ = 22 cos 25 = 19.94 m / s

v_{oy} = 22 sin 25 = 0.0192 m / s

let's use movement on the vertical axis

y = y₀ + v_{oy} t - ½ g t²

when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m

0 = 21 + 0.0192 t - ½ 9.81 t²

4.905 t² - 0.0192 t - 21 = 0

t² - 0.003914 t - 4.2813 =0

we solve the quadratic equation

t = $\frac{ 0.003914\ \pm \sqrt{0.003914^2 + 4 \ 4.2813 } }{2}$

t = $\frac{0.003914 \ \pm 4.13828}{2}$

t₁ = 2.07 s

t₂ = -2.067 s

since time must be a positive scalar quantity, the correct result is

t = 2.07 s

now we can look up the distance traveled

x = v₀ₓ t

x = 19.94 2.07

x = 41.28 m

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Explanation:

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