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A 2.7 kg ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 45.0-m-high cliff. At the instant the ball is thrown, a woman starts running away from the base of the cliff with a constant speed of 6.00 m/s. The woman runs in a straight line on level ground. Ignore air resistance on the ball.A 2.7 kg ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 45.0-m-high cliff. At the instant the ball is thrown, a woman starts running away from the base of the cliff with a constant speed of 6.00 m/s. The woman runs in a straight line on level ground. Ignore air resistance on the ball. How far does the woman run before she catches the ball?

Respuesta :

Answer:

a) 72.54°

b) It will take the woman 33.4m

Explanation:

The woman runs with a constant speed V1=6m/s

V2= 20m/s

V1= V2cos theta

Cos theta= V1/V2= 6/20

Cos theta= 0.3

Cos^-1 0.3=72.54°

b) Using Range formular for projectile

R= (V2Costheta)/g (V2Sintheta)^2 +sqrt(V2Sintheta)^2 + 2gh)

R= (20cos72.54)(2Sin72.54+sqrt(20Sin72.54)^2 + 2×9.8×45

R=33.4m

a) 72.54°

b) It will take the woman 33.4m

  • The calculation is as follows:

(a)

The woman runs with a constant speed V1=6m/s

V2= 20m/s

V1= V2cos theta

Now

Cos theta[tex]= V1\div V2= 6\div 20[/tex]

Cos theta= 0.3

Cos^-1 0.3=72.54°

b)

[tex]R= (V2Cos\theta)\div g (V2Sin\theta)^2 +\sqrt(V2Sin\theta)^2 + 2gh)\\\\R= (20cos72.54)(2Sin72.54+\sqrt(20Sin72.54)^2 + 2\times9.8\times 45[/tex]

R=33.4m

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