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a student drops a ball off the top of building and records that the ball takes 3.32s to reach the ground (g=9.8 m/s^2). what is the ball's speed just before hitting the ground?​

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Luv2Teach

Answer:

Explanation:

Use the one-dimensional equation for motion

v = v₀ + at and filling in,

v = 0 + (-9.8)(3.32) so

v = -33 m/s (negative because it is going downwards and upwards is positive).

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