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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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Answer:

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2 years ago

In billiards, the 0.165 kg cue ball is hit toward the 0.155 kg eight ball, which is stationary. The cue ball travels at 5.8 m/s

Damm [24]

Answer:

another ball velocity = 3.92 m/s and with 30° clockwise from initial direction

Explanation:

given data

mass m1 = 0.165 kg

mass m2 = 0.155 kg

before collision velocity v1 = 5.8 m/s

before collision velocity v2 = 0

angle = 35.0° from initial direction

after collision 1st ball velocity v3 = 3.2 m/s

to find out

after collision another ball velocity v4

solution

we consider here ball move in x axis and after collision 1st ball move upside of x axis with angle 35 degree and other ball move downside with x axis with angle θ

so from conservation of momentum we say

m1v1 = m1v3cos35 + m2v4cosθ with x axis .............1

m1v3sin35 = m2v4sinθ with y axis .............2

so from 1 equation

0.165 × 5.8 = 0.165(3.2)cos35 + 0.155(v4)cosθ

v4 cosθ = 3.38 .................3

form 2 equation

0.165(3.2)sin35 = 0.155(v4)sinθ

v4 sinθ = 1.95 ......................4

so magnitude of another ball velocity is square and adding equation 3 and 4

another ball velocity = √(3.39²+1.96²)

another ball velocity = 3.92 m/s