Question

Young Jeffrey is bored, and decides to throw some things out of the window for fun. But since he is also very curious, he decides to do so in a controlled way. Jeffery takes a rock, his tennis ball, and one of his fathers golf balls and sets them on the sill of an open window. Then he uses a flat board to carefully push them all out of the window at the same time. The 2nd story window is 10.0 ft above the ground, and all three objects hit the ground just 0.500 ft away from the base of the house at the same time, 0.788s after they're dropped. The rock hits with a thud and stays put, but the other two objects bounce. After another 0.591s, the tennis ball reaches a maximum height of 5.63 ft, while the golf ball is 7.88 ft off the ground and still rising. Both have travelled an additional 0.375ft horizontally. Calculate the magnitude of the average velocity of each object during these 1.38 seconds. Which object has the largest instantaneous velocity at the end of this time period?

Answer 1

Answer:

a) Stone            v_{y} = - 7.25 ft / s ,  vₓ = 0.362 ft / s

b) tennis ball    v_{y} =  -3.16 ft / s ,   vₓ = 0.634 ft / s

c) golf ball        v_{y} = - 1,536 ft / s, vx = 0.634 ft / s

2) golf ball

Explanation:

1) The average speed is defined with the displacement interval in the given time interval

           v =( $x_{f}$-x₀) / Δt

let's use this expression for each object

a) Stone

  It tells us that it is released from y₀ = 10 ft and reaches the floor at

t = 0.788 s, but in the problem they tell us that the calculation is for t = 1.38 s

           $v_{y}$ = (0-10) / 1.38

           v_{y} = - 7.25 ft / s

 in this interval a distance of x_{f} = 0.500 ft was moved away from the building (x₀ = 0 ft)

          vₓ = (0.500- 0) / 1.38

          vₓ = 0.362 ft / s

In my opinion it makes no sense to keep measuring the time after the stone has stopped.

b) tennis ball

It leaves the building at a height of y₀= 10ft and at the end of the period it is at a height of y_{f} = 5.63 ft, all this in a time of t = 0.788 + 0.591 = 1.38 s

       

the average vertical speed is

            $v_{y}$ = (5.63 - 10) / 1.38

            v_{y} = -3.16 ft / s

for horizontal velocity the ball leaves the building xo = 0 reaches the floor

x₁ = 0.500 foot and when bouncing it travels x₂ = 0.375 foot, therefore the distance traveled

         x_{f} = x₁ + x₂

         x_{f} = 0.500 + 0.375

         x_{f} = 0.875 ft

we calculate

         vₓ = (0.875 - 0) / 1.38

         vₓ = 0.634 ft / s

c) The golf ball

the vertical displacement y₀ = 10 ft, and y_{f} = 7.88 ft

          v_{y} = (7.88 - 10) / 1.38

          v_{y} = - 1,536 ft / s

the horizontal displacement x₀ = 0 ft to the point xf = 0.875 ft

          vₓ = (0.875 -0) / 1.38

          vₓ = 0.634 ft / s

2) in this part we are asked for the instantaneous speed at the end of the time interval

a) the stone is stopped so its speed is zero

          v_{y} = vₓ = 0

b) the tennis ball

It is at its maximum height so its vertical speed is zero

        v_{y} = 0

horizontal speed does not change

          vₓ = 0.634 ft / s

c) The golf ball

they do not indicate that it is still rising. Therefore its vertical speed is greater than zero

         v_{y} > 0

horizontal speed is constant

         vₓx = 0.634 ft / s

the total velocity of the object can be found with the Pythagorean theorem

         v = √ (vₓ² + v_{y}²)

When reviewing the results, the golf ball is the one with the highest instantaneous speed at the end of the period