Find the magnitude of the velocity vector on impact

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Homework Help Overview

The discussion revolves around calculating the magnitude of the velocity vector of a vehicle at the moment of impact after driving off a cliff. The context involves projectile motion, specifically analyzing the vertical and horizontal components of velocity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different methods to find the final velocity, including the use of kinematic equations and the relationship between horizontal and vertical components of motion. Questions arise regarding the necessity of time in the calculations and the interpretation of specific variables in the equations.

Discussion Status

Some participants have provided alternative approaches to the problem, suggesting that time may not be necessary for the calculations. There is ongoing exploration of the relationships between the components of velocity and the application of kinematic equations, with no explicit consensus reached yet.

Contextual Notes

Participants are working under the constraint of not including units in their answers and are required to provide precision to tenths. The discussion also includes assumptions about neglecting air resistance.

ilovedeathcab
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(resolved) find the magnitude of the velocity vector on impact

Thelma and Louise drive their '66 T-Bird convertible off a cliff. They are going 25.9 m/s and the cliff is 97 meters high. Find the magnitude of their velocity vector the moment they impact the ground. Make sure you do the following in your answer:

DO NOT include units. We'll deal with that in the next question.
Give precision to tenths (#.#).
Make your answer correct to within 0.2

so given y is -97, and vi is 25.9 m/s


equations for horizontal projectile motion for vf are
vyf=-g(t)
vf=sqrt of (vxf^2+vyf^2)
so far for t i had 4.447seconds

____
\/-2*-97
----------= 4.447
9.81
and had vyf as -44.6m/s
 
Last edited:
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You don't need the time.
You can use vf2 = vi2 + 2as for the vertical velocity
Since you ignore air resistance there is no change in the horizontal velocity.
 
And think of vxf and vyf as being two legs of a right triangle. The answer you're looking for is the hypotenuse of this triangle.
 
mgb_phys said:
You don't need the time.
You can use vf2 = vi2 + 2as for the vertical velocity
Since you ignore air resistance there is no change in the horizontal velocity.
what is 2as?
 
2 * acceleration * distance
It's one of the standard motion equations - although some books use different letters.
 

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