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pantera1 [17]
2 years ago
11

While studying projectile motion, we consider ideal scenarios, where the projectile travels along its trajectory only under the

influence of gravity. In real-world situations, however, other forces act on the projectile.
Consider a cargo plane that is dropping relief packages to flood victims. In predicting and studying this motion, we might consider gravity, but ignore the horizontal and vertical forces associated with drag (or air friction). Discuss this simplification. Specifically address these questions:
•Why do we often ignore drag in projectile predictions?
•What conditions (of the object, its surroundings, and its launch) do you think might make drag a significant factor in the relief package drop?
•How would drag affect the projectile's motion if it really were a significant factor in the relief package drop?
Physics
2 answers:
Law Incorporation [45]2 years ago
4 0
Drag is usually ignored because its effect on the horizontal velocity is usually negligible due to the short time of flight.

An object's surface area and geometry, along with the object's surrounding wind speed will affect the drag force.

In most cases, drag force will cause the object to land horizontally closer to the predicted landing point as drag is a resistive force.
cestrela7 [59]2 years ago
3 0

1) The drag can or cannot have an affect on the projectile and has to many variations, such as wind and shape. We use ideal calculations instead.


2) Some conditions that could make drag significant would be mass, shape, and wind. They could all change the spot in which the package lands.


3) If there was something going on such as high winds or a change in general weather the drag would affect the projectile and where it lands.

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We know that acceleration is change in velocity by time taken for that change.

In this case velocity change is 3.7 m/s

Time taken for this change = 60 ms = 6 *10^{-3} seconds

So acceleration of frog  = \frac{3.7}{60*10^{-3}}

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8 0
2 years ago
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A bicyclist of mass 68 kg rides in a circle at a speed of 3.9 m/s. If the radius of the circle is 6.5 m, what is the centripetal
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Data:
Centripetal Force = ? (Newton)
m (mass) = 68 Kg
s (speed) = 3.9 m/s
R (radius) = 6.5 m

Formula:
F_{centripetal\:force} =  \frac{m*s^2}{R}

Solving:
F_{centripetal\:force} = \frac{m*s^2}{R}
F_{centripetal\:force} = \frac{68*3.9^2}{6.5}
F_{centripetal\:force} = \frac{68*15.21}{6.5}
F_{centripetal\:force} = \frac{1034.28}{6.5}
\boxed{\boxed{F_{centripetal\:force} = 159.12\:N}}
Answer:
<span>B.159 N</span>
3 0
2 years ago
Problem: An ice hockey player hits a puck of mass 0.15 kilograms with a force of 100 newtons in the horizontal direction. What i
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Substituting the values,
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In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. controlling the angle of the sp
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V ( initial ) = 20 m/s
h = 2.30 m
h = v y * t + g t ² / 2
d = v x * t
1 ) At α = 18°:
v y = 20 * sin 18° = 6.18 m/s
v x = 20 * cos 18° = 19.02 m/ s
2.30 = 6.18 t + 4.9 t²
4.9 t² + 6.18 t - 2.30 = 0
After solving the quadratic equation ( a = 4.9, b = 6.18, c = - 2.3 ):
t 1/2 = (- 6.18 +/- √( 6.18² - 4 * 4.9 * (-2.3)) ) / ( 2 * 4.9 )  
t = 0.3 s
d 1 = 19.02 m/s * 0.3 s = 5.706 m
2 ) At  α = 8°:
v y = 20* sin 8° = 2.78 m/s
v x = 20* cos 8° = 19.81 m/s
2.3 = 2.78 t + 4.9 t² 
4.9 t² + 2.78 t - 2.3 = 0
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d 2 = 19.81 * 0.46 = 9.113 m
The distance is:
d 2 - d 1 = 9.113 m - 5.706 m = 3.407 m

GOOD LUCK AND HOPE IT HELPS U
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