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?
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Ответ:
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
Ответ: