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3 years ago

9

A ball is shot straight up into the air from the ground with initial velocity of 45ft/sec. Assuming that the air resistance can

be ignored, how high does it go?

1 answer:

Nesterboy [21]3 years ago

6 0

V2^2 = V1^2 - 2gh

<span>=> 0 = 44^2 - 2 (32.2) h </span>

<span>=> h = 44^2 / 64.4 = 30.062 ft</span>

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