Calculating Time to Apex for Vertically Launched Objects

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To calculate the time it takes for a vertically launched object to reach its apex, the final velocity at the highest point is zero. Using the equation v(final) = Vo + a(t), where Vo is the initial velocity (29.4 m/s) and acceleration (a) is -9.8 m/s², the equation becomes 0 m/s = 29.4 m/s - 9.8t. Rearranging gives -29.4 m/s = -9.8t, leading to t = 3 seconds to reach the apex. The discussion confirms that the object returns to the same speed when coming down, but the time to reach the highest point is correctly calculated as 3 seconds.
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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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