g Suppose that an escalator raises stationary passengers vertically at a speed of ten feet per second, and the escalator is pointing northeast so its passengers also move northeast at ten feet per second. If a child riding the escalator drops a ball so that it rolls north on the escalator at a speed of 5 feet per second, what is the speed of the ball relative to a stationary observer

v = 14 ft / s, 49.6º north of east

Explanation:

This is a relative velocity problem, let's use subscripts to work

* e for the escalator

* p for ball

velocity is a vector whereby

     v = v_e + v_p

The bold are vector.  An easy way to work with this type of problem is to decompose the velocities into a Cartesian coordinate system, where the x-axis coincides with East and the y-axis with North.

stair speed

              sin 45 = v_{ey} / ve

              cos 45 = v_{ex} / ve

              v_{ey} = go sin 45

               v_{ex} = ve cos 45

              v_{ey} = 10 sin 45 = 7.07 ft / s

               v_{ex} = 10 cos 45 = 7.07 ft / s

the speed of the ball is

                v_p = 5 j ^ ft / s

therefore the speed that the observer sees is

X axis

            vₓ = v_{ex}

             vₓ = 7.07 ft / s

Y axis  

            v_y = v_{ey} + v_p

            v_y = 7.07 + 5

            v_y = 12.07 ft / s

we can give the result in two ways

*      v  = (7.07 i ^ +12.07 j ^) ft / s

* in module and angle form

Let's use the Pythagorean theorem

         v =

         v =

         v = 13.99 ft / s

         v = 14 ft / s

we use trigonometry

        tan θ = vy / vx

        θ = tan⁻¹ v_y / vₓ

        θ = tan⁻¹ 12.07 / 7.07

        θ = 49.6º

in the form of cardinal coordinates they are 49.6º north of east

momentum of soft ball three times less than baskey ball

more than, less

less than, more