A hiker throws a rock from ground level at 11.3 m/s at an angle of 26 degrees over a river that is 6.27 m wide. What height is the rock at when it reaches the other side of the river? (Unit = m) Watch sig figs.

y = 1.19 m

Explanation:

For this exercise, let's use the projectile launch ratios, find the components of the initial velocity

          sin 26 = v_{oy} / v₀

          cos 26 = v₀ₓ / v₀

          v_{oy} = v₀ sin  26

          v₀ₓ = v₀ cos 26

          v_{oy} = 11.3 sin 26 = 4.95 m / s

          v₀ₓ = 11.3 cos 26 = 10.16 m / s

Let's find the time it takes to cross the river

            x = v₀ₓ t

            t = x / v₀ₓ

            t = 6.27 / 10.16

            t = 0.617 s

now with that time we can find how high the body is

           y = y₀ + v_{oy} t - ½ g t²

let's calculate

           y = 0 + 4.95 0.617 - ½ 9.8 0.617²

           y = 1,189 m

the result must have three significant figures

           y = 1.19 m