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Will the change in kinetic energy i.e. $\frac{1}{2}mv^2-\frac{1}{2}mu^2$ always zero in case of projectile motion since the initial speed and final speed is the same that is $u$? If my observation is wrong please correct me with the help of diagrams.

Please also say me that will the change in K.E. be zero when the object is projected from a certain height above ground both horizontally and also when making angle $\theta$ with x axis

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  • $\begingroup$ The projectile may land at a different elevation than it was launched from. What does that tell you? $\endgroup$ Sep 12 at 15:59
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Assuming you are talking about projectile motion in the presence of a uniform gravitational field the answer is no.

The total energy of a particle of mass $m$ is given by $E=\frac{1}{2}mv^2 +mgy $ where $y$ is the height of the projectile from the floor and $v^2$ is the magnitude of the velocity squared.

As the projectile launches up kinetic energy is converted to gravitational potential energy. The total energy remains the same so the difference in kinetic energy at any given time is just $\Delta E= mgy(t)$.

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