Question

You are standing on a bridge that is 12 meters above the water. If you throw a rock with a speed of 12 m/s at an angle of 20 degrees above the horizontal, how far will it travel horizontally when it hits the water?

Answer 1

x = 22.99 m

Step-by-step explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

sin 20 =
`v_(oy)` / v₀

cos 20 = v₀ₓ / v₀

v_{oy} = v₀ sin 20

v₀ₓ = v₀ cos 20

v_{oy} = 12 sin 20

v₀ₓ = 12 cos 20

v_{oy} = 4.1 m / s

v₀ₓ = 11.28 m / s

let's look at the time it takes to reach the ground

y = y₀ + v_{oy} t - ½ g t²

where i is the height of the bridge 12m

0 = 12 + 4.1 t - ½ 9.8 t²

t² - 0.837 t - 2.45 = 0

let's solve the quadratic equation

t = [0.837 ±√ (0.837² +4 2.45)] / 2

t = [0.837 ± 3.24] / 2

t₁ = -1.2 s

t₂ = 2.0385 s

as the time has to be positive the correct result is t = 2.04 s

we use this time to calculate the horizontal distance traveled

x = v₀ₓ t

x = 11.28 2.0385

x = 22.99 m