An object fired upward at a speed of 3.000 m/s

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The discussion centers on the physics of an object fired upward at an initial speed of 3.000 m/s, with gravitational acceleration set at g = -9.81 m/s². The velocity of the object at its maximum height is definitively 0 m/s, as it momentarily stops before descending. To calculate the maximum height reached, participants suggest using the kinematic equation V_f² = V_i² + 2ad, where V_f is the final velocity (0 m/s), V_i is the initial velocity (3.000 m/s), and a is the acceleration due to gravity (-9.81 m/s²).

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Diff. ways possible for velocity of an object to be changing when speed is not!

just a fun question? It would be nice if u could answer it! I also have questions with the problem below!



can you help me w/ the rest of this problem.

at t= 0.0000 s. It reaches its max. height at time t. What is its velocity at the max. height? Assuming that g=(-9.81 m/s/s), calculate the max. height reached in meters:

Any kind of help is appreciated! :smile:
 
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SpeedeRi2 said:
just a fun question? It would be nice if u could answer it! I also have questions with the problem below!



can you help me w/ the rest of this problem.

at t= 0.0000 s. It reaches its max. height at time t. What is its velocity at the max. height? Assuming that g=(-9.81 m/s/s), calculate the max. height reached in meters:

Any kind of help is appreciated! :smile:
You can do this using [tex]\Delta K + \Delta U = 0[/tex]

or you can use the kinematic equations to solve this

for [tex]\Delta K + \Delta U = 0[/tex]
[tex]\frac{1}{2} m v_{i} = mgh_{2}[/tex]

for the kinematic equations use
[tex]d = v_{1} t + \frac{1}{2} a t^2[/tex]
 
the velocity at its maximum point would be 0 because it is about to start falling back down.
using the formula [tex]V_{f}^{2} = V_{i}^{2} + 2ad[/tex], you can get the distance it travels upwards.
[tex]V_{f} =[/tex]0
[tex]V_{i} =[/tex]3
a = -9.8
Solve for d.
 

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