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3 years ago

A foul ball is hit straight up into the air with a speed of about 25 m/s. ? how high does it go? (b) how long is it in the air?

Physics

1 answer:

nikdorinn [45]3 years ago

5 0

To solve this problem, we make use of the equations of motion:

vf^2 = v0^2 – 2 g h

vf = v0 – g t

h = v0 t – 0.5 g t^2

A. Finding for max height:

Max height occurs when vf = 0, therefore:

0^2 = 25^2 – 2 (9.8) h

h = 31.89 m

B. Finding for t:

t = t(up) + t(down)

vf = v0 – g t(up)

0 = 25 – 9.8 t(up)

t(up) = 2.55 s

h = v0 t(down) – 0.5 g t(down)^2

31.89 = 0.5 (9.8) t(down)^2

t(down)^2 = 6.51

t(down) = 2.55 s

so,

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We use the equation of motion for vertical component,

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Here, $s_{y}$ is displacement of bullet, $u_{y}$ is vertical initial velocity of bullet which is equal to zero because bullet was fired horizontally, and t is time of flight.

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Substituting the values, we get time of flight

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Answer:

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