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3 years ago

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An Alaskan rescue plane traveling 41 m/s drops a package of emergency rations from a height of 192 m to a stranded party of expl

orers. The acceleration of gravity is 9.8 m/s 2 . Where does the package strike the ground relative to the point directly below where it was released? Answer in units of m. 008 (part 2 of 3) 10.0 points What is the horizontal component of the velocity just before it hits? Answer in units of m/s. 009 (part 3 of 3) 10.0 points What is the vertical component of the velocity just before it hits? (Choose upward as the positive vertical direction) Answer in units of m/s.

1 answer:

svet-max [94.6K]3 years ago

4 0

Answer:

a)The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) Vertical component of velocity = 61.41 m/s

Explanation:

a) Consider the vertical motion of plane,

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Displacement, s = 192 m

Acceleration, a = 9.81 m/s²

Substituting

s = ut + 0.5 at²

192 = 0 x t + 0.5 x 9.81 x t²

t = 6.26 seconds

Now we need to find horizontal distance traveled in 6.26 seconds by the package.

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 41 m/s

Time, t = 6.26 s

Acceleration, a = 0 m/s²

Substituting

s = ut + 0.5 at²

s = 41 x 6.26 + 0.5 x 0 x 6.26²

s = 256.52 m

The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) We have equation of motion, v = u+ at

Initial velocity, u = 0 m/s

Time, t = 6.26 s

Acceleration, a = 9.81 m/s²

Substituting

v = u+ at

v = 0 + 9.81 x 6.26 = 61.41 m/s

Vertical component of velocity = 61.41 m/s

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Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 27.0°below the horizontal. If it st

andreev551 [17]

a) The time of flight is 3.78 s

b) The initial speed is 17.0 m/s

c) The speed at impact is 46.4 m/s at $70.3^{\circ}$ below the horizontal

Explanation:

The picture of the previous problem (and some data) is missing: find it in attachment.

a)

The motion of the ball is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

We start by considering the vertical motion, to find the time of flight of the ball. We do it by using the following suvat equation: for the y-displacement:

$y=u_y t+\frac{1}{2}at^2$

where we have:

y = -45.0 m is the vertical displacement of the ball (the height of the building)

$u_y=u sin \theta$ is the initial vertical velocity, with u being the initial velocity (unknown) and $\theta=-27.0^{\circ}$ the angle of projection

t is the time of the fall

$a=g=-9.8 m/s^2$ is the acceleration of gravity

Along the x-direction, the equation of motion is instead

$x=(u cos \theta)t$

where $ucos \theta$ is the horizontal component of the velocity. Rewriting this equation as

$t=\frac{x}{ucos \theta}$

And substituting into the previous equation, we get

$y=xtan \theta + \frac{1}{2}gt^2$

And using the fact that the horizontal range is

x = 59.0 m

And solving for t, we find the time of flight:

$t=\sqrt{\frac{y-x tan \theta}{g}}=\sqrt{\frac{-45-(59.0)(tan(-27^{\circ}))}{-9.8}}=3.78 s$

b)

We can now find the initial speed, u, by using the equation of motion along the x-direction

$x=u cos \theta t$

where we know that:

x = 59.0 m is the horizontal range

$\theta=-23^{\circ}$ is the angle of projection

$t=3.78 s$ is the time of flight

Solving for u, we find the initial speed:

$u=\frac{x}{cos \theta t}=\frac{59.0}{(cos (-23^{\circ}))(3.78)}=17.0 m/s$

c)

First of all, we notice that the horizontal component of the velocity remains constant during the motion, and it is equal to

$v_x = u cos \theta = (17.0)(cos (-23^{\circ})=15.6 m/s$

The vertical velocity instead changes according to the equation

$v_y = u sin \theta + gt$

Substituting all the values and t = 3.78 s, the time of flight, we find the vertical velocity at the time of impact:

$v_y = (17.0)(sin (-23^{\circ}))+(-9.8)(3.78)=-43.7 m/s$

Where the negative sign means it is downward.

Therefore, the speed at impact is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(15.6)^2+(-43.7)^2}=46.4 m/s$

while the direction is given by

$\theta = tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-43.7}{15.6})=-70.3^{\circ}$

So, $70.3^{\circ}$ below the horizontal.

Learn more about projectile motion:

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#LearnwithBrainly

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3 years ago

A new system of units is invented where the basic unit of length is the "dak." The "dak" is defined as equal to 3 inches, and th

Goryan [66]

Answer:

1 cm = 0.131 daks

Explanation:

It is given that,

1 dak = 3 inches

and

1 inch = 2.54 cm

We need to find how many "daks" are there in a centimeter. We can do the conversions as follows :

$1\ dak=1\ dak\times \dfrac{3\ inches}{1\ dak}\times \dfrac{2.54\ cm}{1\ inch}$

$1\ dak=7.62\ cm$

$1\ cm=\dfrac{1}{7.62}\ dak$

1 cm = 0.131 dak

So, there are 0.131 daks in 1 centimeter. Hence, this is the required solution.

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4 years ago

Explain the differences between horticulture and agriculture.

sammy [17]

<h2><u>Hello </u><u>There</u></h2>

$\tt{Difference \: Between \: \underline\red {Horticulture} \: And \: \underline\green{Agriculture}}$

$\text{\underline\red {Horticulture}}$

The Cultivation of Fruits, Vegetables and Flowers for domestic and international markets are called Horticulture.

$\text{\underline\green{Agriculture}}$

It is science, art and occupation of cultivating the soil, producing crops and livestock. It is systematic and controlled use of living organisms and the environment to improve the human condition.

<h3>Hope This Helps</h3>

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2 years ago

A projectile is launched with an initial velocity of (40 m/s), at an angle of (30°) above

zlopas [31]

Answer:

330.5 m

Explanation:

In this case, the object is launched horizontally at 30° with an initial velocity of 40 m/s .

The maximum height will be calculated as;

$h=\frac{v^2_isin^2\alpha }{2g}$

where ∝ is the angle of launch = 30°

vi= initial launch velocity = 40 m/s

g= 10 m/s²

h= 40²*sin²40° / 2*10

h={1600*0.4132 }/ 20

h= 661.1/2 = 330.5 m

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3 years ago

A 65 Kg roller-blade is accelerating at 5 m/s/s across the side walk. What force would be necessary for this acceleration to occ

Neporo4naja [7]

Newton's second law states that the force applied to an object is equal to the product between the mass m of the object and its acceleration a:
$F=ma$
Using $m=65 kg$ and $a=5 m/s^2$ , we can find the value of the force applied to the roller-blade to obtain this acceleration:
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3 years ago