The Physics of a Stunt Man's Cliff Jump: Solving for Time, Speed, and Distance

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SUMMARY

The discussion focuses on solving a physics problem involving a stunt man's cliff jump, where he runs horizontally off a cliff at a speed of 6.0 m/s and lands in a river 23 meters below. Key calculations include determining the time of flight, horizontal and vertical components of speed upon impact, total speed at impact, and the angle of entry into the water. The equations of motion are utilized, specifically D=(V2)(t)-0.5(a)(t²) for distance and components of velocity using trigonometric functions.

PREREQUISITES
  • Understanding of basic physics concepts, particularly projectile motion
  • Familiarity with kinematic equations and their applications
  • Knowledge of trigonometric functions and their role in resolving vector components
  • Ability to perform algebraic manipulations to isolate variables in equations
NEXT STEPS
  • Study the principles of projectile motion in physics
  • Learn how to apply kinematic equations to solve for time, speed, and distance
  • Explore the use of trigonometric functions in resolving velocity components
  • Practice solving similar physics problems involving vertical and horizontal motion
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of projectile motion and its applications in real-world scenarios, such as stunt performance in film.

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Basics Physics...Need Help

Homework Statement



-In a movie, a stunt man runs horizontally off a cliff at a speed of 6.0 m/s. The stunt man lands in a deep river that is 23m below the cliff.


A) Determine when the stunt man lands in the river.

B) Determine the horizontal component of the stunt man's speed as he hits the river.

C) Determine the vertical component of the stunt man's speed as he hits the river.

D) Determine the stunt man's speed as he hits the river.

E) Determine the angle he hits the water at.

F) How far from the base of the cliff does the stunt man hit the water ?



Homework Equations



V1y=v1sin (feta=angle)

V1x=V1cos (feta)

D=(V2)(t)-0.5(a)(t-squared)

The Attempt at a Solution



A) D=(V2)(t)-0.5(a)(t-squared)
re-arranges to d+0.5(a) = t ?
V2

B) ?


i have no clue my teacher is terrible ?
 
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first of all, don't change variables in the middle of a solution ...
your equations for v1x and v1y are okay , but what is v2 ?

what quantity do you think "D" is ? ... is "d" the same quantity?

second, you have to use proper algebra to re-arrange (to isolate t^2 ...)
 

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