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A table-tennis ball is released near the surface of the (airless) moon with velocity v0 = (0, 5, −3) m/s. It accelerates (downward) with acceleration a = (0, 0, −1.6) m/s2. Find its velocity after 5 s.

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u = 5\(\hat i-3\hat k\)

t = 5

a = -1.6 \(\hat k\)

using equation

v = u + at

v = 5\(\hat j\) - 3\(\hat k\) + (-1.6\(\hat k\))5

v = 5\(\hat j\) - 3\(\hat k\) - 8\(\hat k\)

v = 5\(\hat j\) - 11\(\hat k\)

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