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 13.0 m/s along a straight line and the vertical distance between the limb and the level of the saddle is 3.22 m. (Define the point at which the cowboy's bottom and the saddle meet as (x, y) = (0,0).)

Required:
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?

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moya1316 moya1316 · Answer 1 · 2021-01-31T13:25:31+01:00

Answer:

x = 10.53 m

Explanation:

Let's analyze this problem a bit, the time that the cowboy must take to fall must be the time that the horse takes to arrive

Let's start by looking for the cowboy's time, which starts from rest and the point where the chair is is y = 0

y =y₀ + v₀ t - ½ g t²

0 = y₀ - ½ g t²

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

we calculate

t = √(2 3.22 / 9.8)

t = 0.81 s

the horse goes at a constant speed

x = [tex]v_{c}[/tex] t

x = 13 0.81

x = 10.53 m

this is the distance where the horse should be when in cowboy it is left Cartesian

ConcepcionPetillo ConcepcionPetillo · Answer 2 · 2022-01-21T13:37:52+01:00

The position of the horse from the cowboy must be 10.53 m apart.

The time taken by the cowboy to fall must be equal to the time that the horse takes to arrive at the point where the cowboy falls.

We can calculate the time taken by the cowboy to reach the saddle by applying the second equation of motion

h = ut + (1/2)gt²

here h =3.22m , u = 0m/s as he starts from rest

[tex]t=\sqrt[]{\frac{2h}{g} }\\ t=\sqrt{\frac{2*3.22}{9.8} } [/tex]

t = 0.81 s

the horse goes at a constant speed

x = vt , given that v = 13 m/s

x = 13 × 0.81

x = 10.53 m is the distance where the horse should be when the cowboy jumps.

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