A10-kg projectile is fired straight up with an initial velocity of 500 m/s. (a) what is the projectile’s potential energy at the top of its trajectory? (b) what would be the maximum potential energy if the projectile were fired at a 450 angle rather than straight up?

(a) the initial kinetic energy of the projectile is equal to:

The projectile is fired straight up, so at the top of its trajectory, its velocity is zero; this means that it has no kinetic energy left, so for the law of conservation of energy, all its energy has converted into potential energy, which is equal to


b) If the projectile is fired with an angle of , its velocity has 2 components, one in the x-direction and one in the y-direction:



This means that at the top of its trajectory, only the vertical velocity will be zero (because the horizontal velocity is constant, since the motion on the x-axis is a uniform motion). Therefore, at the top of the trajectory, the projectile will have some kinetic energy left:

For the conservation of energy, the initial energy mechanical energy must be equal to the mechanical energy at the highest point:

the initial kinetic energy is the same as point a), so we can re-arrange this equation to find the new potential energy at the top of the trajectory:

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