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Basile [38]
4 years ago
11

What effect does air resistance have on the projectile

Physics
2 answers:
Vesnalui [34]4 years ago
7 0

more deceleration.

in vertical motion downwards => terminal velocity ... raindrops etc

tino4ka555 [31]4 years ago
7 0

Answer:

Decreases its maximum acceleration

Explanation:

Air resistance which is also known as drag force occurs with an object in air while in motion, it encounters air resistance which reduces its speed and reduces its maximum height. Example of bodies in which air resistance affects are rockets, bullets, a ball in air etc

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Alex17521 [72]
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4 years ago
A fish swimming in a horizontal plane has velocity i = (4.00 î + 1.00 ĵ) m/s at a point in the ocean where the position relative
Liula [17]

Question is missing. Found on google:

a) What are the components of the acceleration of the fish?  

(b) What is the direction of its acceleration with respect to unit vector î?

(c) If the fish maintains constant acceleration, where is it at t = 30.0 s?

(a) (0.73, -0.47) m/s^2

The initial velocity of the fish is

u=(4.00 i + 1.00 j) m/s

while the final velocity is

v=(15.0 i - 6.00 j) m/s

Initial and final velocity are related by the following suvat equation:

v=u+at

where

a is the acceleration

t is the time

The time in this case is t = 15.0 s, so we can use the previous equation to find the acceleration, separating the components:

v_x = u_x + a_x t\\a_x = \frac{v_x-u_x}{t}=\frac{15.0-4.00}{15.0}=0.73 m/s^2

v_y = u_y + a_y t\\a_y = \frac{v_y-u_y}{t}=\frac{-6.00-1.00}{15.0}=-0.47 m/s^2

(b) -32.8^{\circ}

The direction of the acceleration vector with respect to i can be found by using the formula

\theta = tan^{-1}(\frac{a_y}{a_x})

where

a_x is the horizontal  component of the acceleration

a_y is the vertical component of the acceleration

From part a), we have

a_x = 0.73 m/s^2

a_y = -0.47 m/s^2

Substituting,

\theta = tan^{-1}(\frac{-0.47}{0.73})=-32.8^{\circ}

(c) r=(460.5 i - 185.1 j )m

The initial position of the fish is

r_0 = (12.0 i -3.60 j) m

The generic position r at time t is given by

r= r_0 + ut + \frac{1}{2}at^2

where

u=(4.00 i + 1.00 j) m/s is the initial velocity

a=(0.73 i -0.47 j) m/s^2 is the acceleration

Substituting t = 30.0 s, we find the final position of the fish. Separating each component:

r_x =12.0 + (4.00)(30) + \frac{1}{2}(0.73)(30)^2=460.5 m\\r_y = -3.60 + (1.00)(30) + \frac{1}{2}(-0.47)(30)^2=-185.1 m

So the final position is

r=(460.5 i - 185.1 j )m

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