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 a speed of 3.000 m/s, exploring how velocity can change even when speed remains constant. Participants are asked to calculate the object's velocity at its maximum height, which is zero, and to determine the maximum height reached using gravitational acceleration of -9.81 m/s². Two methods are suggested: using the principle of conservation of energy or applying kinematic equations. The kinematic equation V_f² = V_i² + 2ad is highlighted for calculating the height, with the initial velocity and acceleration provided. The conversation emphasizes understanding the relationship between speed, velocity, and height in projectile motion.
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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 \Delta K + \Delta U = 0

or you can use the kinematic equations to solve this

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

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