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You are working in the movie industry. The director of a western film has come up with an impressive stunt and has asked you to assist with its setup. A cowboy sitting on a tree limb is to drop vertically from rest onto his galloping horse as it passes under the limb. The horse gallops at a constant rate of 11.0 m/s along a straight line and the vertical distance between the limb and the level of the saddle is 3.57 m. (Define the point at which the cowboy's bottom and the saddle meet as (x, y) = (0,0).)
(a) You must advise the director as to the position of the horse along the line of its travel when the cowboy should begin his drop. What advice do you provide the director? (Give the magnitude of your answer in m.) m
(b) The director also asks you for advice on how much padding to put on the saddle so that the cowboy is not injured. (Assume the cowboy's flesh is compressed 3.0 cm against the saddle before coming to rest.)
The cowboy would avoid injury if the director adds at least 42.7 cm of padding to the saddle.
The amount of padding necessary to avoid injury would be too thick for the director to hide from the film's audience.
The cowboy would avoid injury if the director adds at least 3.0 cm of padding to the saddle.
No additional padding is necessary and the cowboy's body would withstand the landing without injury.

Sagot :

what grade is this if lower than collage i can help

kinematics  were used able to find the answers

a) the horse must be at x = 9.39 m when the cowboy lets go

b) the correct answer about the padding is:

The cowboy would avoid injury if the director adds at least 42.7 cm of padding to the saddle., The exact value is y = 32.7 cm

given parameters

a)  the horse's speed [tex]v_{c}[/tex] = 11.0 m / s

    cowboy's initial height y₀ = 3.57 m

b)  padding thickness

    distance compressing the meat y = 3 cm (1 m / 100cm) = 0.03 m

to find

a ) the horizontal distance to the horse

b)  need for padding

After reading this extensive problem we are going to solve it in parts

a) Let's use the laws of kinematics to find the time it takes the cowboy to reach the horse saddle, since his fall from rest begins, his initial velocity is zero

          y = y₀ + v₀ t - ½ g t²

where y₀ is the initial height, and the height when it reaches the saddle (y = 0), v₀ the initial velocity, g the acceleration of gravity (g = 9.80 m / s²), t the time

           0 = y₀ + 0 - ½ g t²

           t = [tex]\sqrt{\frac{2y_o}{g} }[/tex]

           t = [tex]\sqrt{\frac{2 \ 3.57}{9.8} }[/tex]

          t = 0.854 s

this is the same time the horse must move, let's use the uniform motion ratios

           [tex]v_c[/tex] = x / t

            x = [tex]v_c[/tex] t

            x = 11 0.854

            x = 9.39 m

Therefore, using the concepts of kinematics, we find the distance that the horse must be for the cowboy to let go is x = 9.39 m

b) For this part we look in tables how much weight the bones of the spine resist and it is a value between 600 kg to 800 kg, therefore the weight that the spine resists

         W = m g

          W₁ = 600 9.8 = 5880 N

          W₂ = 800 9.8 = 7840 N

To answer this part we use Newton's second law to find the maximum strength that the cowboy's column can resist, which is the maximum force (W2)

           F = m a

           a = 7840 / m

Let's start using kinematics to find the cowboy speed when he reaches the saddle.

           v₁ = v₀ - g t

where v₁ is the velocity at the end of the interval, v₀ the initial velocity (v₀=0), g the acceleration of gravity and t the time

          v₁ = 0 - g t

          v₁ = - 9.8 0.854

          v₁ = - 8.369 m / s

the negative sign indicates that the speed is down

now let's look for braking acceleration

        v² = v₁² - 2 a y

when the cowboy stops his velocity is zero (v = 0)

       0 = v₁² - 2 a y

        y = [tex]\frac{v_1^2}{2a }[/tex]

        y = [tex]\frac{8.369^2 \ m}{2 \ 7840}[/tex]

        y = 0.00447 m

In order to complete the calculation, the cowboy mass is needed, suppose it is an average person

        m = 80 kg

         y = 0.00447   80

         y = 0.357 m

         y = 35.7 cm

This is the distance that the system must be compressed so that the force on the column does not exceed its breaking point. The thickness of the system is composed of the compression of the cowboy's meat plus the padding of the saddle

         y = 3 + [tex]e_{chair}[/tex]r

         [tex]e_{chair}[/tex] = 35.7 - 3

         [tex]e_{chair}[/tex] = 32.7 cm

Therefore, withthe Newton's second law and  kinematics, finding it the correct answer is

  •  The cowboy would avoid injury if the director adds at least 42.7 cm of padding to the saddle.

learn more about Newton's second law and kinematics here : brainly.com/question/19315467

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