what is the acceleration and velocity of a projectile at the top of it's trajectory
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So the acceleration of the projectile is equal to the acceleration due to gravity, 9.81 meters/second/second, from just after its thrown, through its highest point, and until just before it hits the ground.
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The highest point in any trajectory, called the apex, is reached when vy = 0. Since we know the initial and final velocities, as well as the initial position, we use the following equation to find y: v2y=v20y−2g(y−y0).
So the acceleration of the projectile is equal to the acceleration due to gravity, 9.81 meters/second/second, from just after its thrown, through its highest point, and until just before it hits the ground.
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