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

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A stunt man runs off a cliff at 6.0 m/s, landing in a river 23m below. To determine the time of flight, the horizontal and vertical components of speed, and the distance from the cliff base, proper physics equations must be applied. The vertical motion can be analyzed using the equation D=(V2)(t)-0.5(a)(t^2), where D is the vertical distance. Clarification on variables and correct algebraic manipulation is essential for solving the problem. Understanding the distinction between variables like "D" and "d" is crucial for accurate calculations.
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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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