Calculating Time to Apex for Vertically Launched Objects

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Homework Help Overview

The discussion revolves around calculating the time it takes for an object launched vertically upwards to reach its apex, with an initial velocity of 29.4 m/s and considering the acceleration due to gravity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of kinematic equations to relate initial velocity, final velocity, acceleration, and time. Questions arise regarding the correct application of these equations, particularly concerning the final velocity at the apex and the implications of the object's return journey.

Discussion Status

Participants have engaged in clarifying the conditions at the apex, with some confirming that the final velocity should be set to zero at that point. There is a recognition of the relationship between upward and downward motion, but the focus remains on the specific question of time to reach the apex.

Contextual Notes

There is an ongoing discussion about the assumptions related to the motion of the object, particularly the effects of gravity and the definition of final velocity at the apex. Some participants express uncertainty about the calculations and interpretations of the kinematic equations.

get_rekd
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how do you figure out how long it take's an object launched vertically upwards to reach its apex?
V=29.4 m/s
 
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v(final)= Vo + a(t) (a= -9.8)
 
so
-29.4 m/s = 29.4 m/s + -9.8t?

-58.8 m/s = -9.8t
----------- ------
-9.8 -9.8

6s = t to reach apex
 
Last edited:
Vo is given
 
no when it reaches its apex (highest point), the ball has a speed of zero
 
yeah so it is v(final)=29.4 m/s + a(t)
but doesn't an object come back down at the same velocity that it went up?
that is why I thought:
-29.4 m/s = 29.4 m/s + a(t)
 
right, that's true. but that's not what the question asks. It asks when it reaches its highest point, at the highest point the velocity is 0
 
okay so then I should set final velocity to = 0 m/s?

0 m/s = 29.4 m/s + -9.8t
-29.4 m/s = -9.8 t
---------- -------
-9.8 -9.8

t = 3 s ?
 
anybody?
 
  • #10
yo, that's right
 
  • #11
thanks
 

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