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

they are playing too much beyblade in the office

explanation:

as we probably am aware that our atmosphere or air have various levels, each level has different gasses, and each gas has another porousness. at the point when the light originates from lower porousness to higher penetrability, it alters its course. with the goal that implies that when the light originates from the moon to the earth, it alters it's course a lot more occasions. it essentially implies that the state of the light piece of the moon is changed on the earth.

not copied too! hope this !

joke:

h o h o hope you have a merry christmas!