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2 years ago

What is the equation to find an angle of projectile

Physics

1 answer:

Ksju [112]2 years ago

3 0

I do not recall the answer to this question

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You raise a bucket of water from the bottom of a deep well. if your power output is 138 w , and the mass of the bucket and the w

inn [45]

Maybe 34.5

I'm not quite sure

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3 years ago

One strategy in a snowball fight is to throw

faltersainse [42]

Answers:

a) $\theta_{2}=23\°$

b) $t=1.199 s$

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the snowball has two components: <u>x-component</u> and <u>y-component</u>. Being their main equations as follows for both snowballs:

<h3><u>Snowball 1:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{1} t_{1}$ (1)

Where:

$V_{o}=11.1 m/s$ is the initial speed of snowball 1 (and snowball 2, as well)

$\theta_{1}=67\°$ is the angle for snowball 1

$t_{1}$ is the time since the snowball 1 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (2)

Where:

$y_{o}=0$ is the initial height of the snowball 1 (assuming that both people are only on the x axis of the frame of reference, therefore the value of the position in the y-component is zero.)

$y=0$ is the final height of the snowball 1

$g=-9.8m/s^{2}$ is the acceleration due gravity (always directed downwards)

<h3><u>Snowball 2:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{2} t_{2}$ (3)

Where:

$\theta_{2}$ is the angle for snowball 2

$t_{2}$ is the time since the snowball 2 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{2} t_{2}+\frac{gt_{2}^{2}}{2}$ (4)

Having this clear, let's begin with the answers:

<h2>a) Angle for snowball 2</h2>

Firstly, we have to isolate $t_{1}$ from (2):

$0=0+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (5)

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$ (6)

Substituting (6) in (1):

$x=V_{o}cos\theta_{1}(-\frac{2V_{o}sin\theta_{1}}{g})$ (7)

Rewritting (7) and knowing $sin(2\theta)=sen\theta cos\theta$ :

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{1})$ (8)

$x=-\frac{(11.1 m/s)^{2}}{-9.8 m/s^{2}} sin(2(67\°))$ (9)

$x=9.043 m$ (10) This is the point at which snowball 1 hits and snowball 2 should hit, too.

With this in mind, we have to isolate $t_{2}$ from (4) and substitute it on (3):

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$ (11)

$x=V_{o}cos\theta_{2} (-\frac{2V_{o}sin\theta_{2}}{g})$ (12)

Rewritting (12):

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{2})$ (13)

Finding $\theta_{2}$ :

$2\theta_{2}=sin^{-1}(\frac{-xg}{V_{o}^{2}})$ (14)

$2\theta_{2}=45.99\°$

$\theta_{2}=22.99\° \approx 23\°$ (15) This is the second angle at which snowball 2 must be thrown. Note this angle is lower than the first angle $(\theta_{2} < \theta_{1})$ .

<h2>b) Time difference between both snowballs</h2>

Now we will find the value of $t_{1}$ and $t_{2}$ from (6) and (11), respectively:

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$

$t_{1}=-\frac{2(11.1 m/s)sin(67\°)}{-9.8m/s^{2}}$ (16)

$t_{1}=2.085 s$ (17)

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$

$t_{2}=-\frac{2(11.1 m/s)sin(23\°)}{-9.8m/s^{2}}$ (18)

$t_{2}=0.885 s$ (19)

Since snowball 1 was thrown before snowball 2, we have:

$t_{1}-t=t_{2}$ (20)

Finding the time difference $t$ between both:

$t=t_{1}-t_{2}$ (21)

$t=2.085 s - 0.885 s$

Finally:

$t=1.199 s$

7 0

2 years ago

A spotlight on the ground shines on a wall 15 meters away. A man 1.8 meters tall walks away from the spotlight toward the buildi

Alex17521 [72]

Answer:

$s=(15-x)m$

Explanation:

Let the distance from spotlight to wall be 15m, and distance from the man to the building be $x$ .

#Therefore the height of the shadow as a function of the above is $s=(15-x )m$

Hence, height of the shadow is expressed as s=(15-x)m

#See attached photo for illustration

6 0

3 years ago

An airplane is initially flying horizontally (not gaining or losing altitude), and heading exactly North. Suppose that the earth

yanalaym [24]

Note: The answer choices are :

a) Increased

b) Decreased

c) stayed the same

Answer:

The correct option is Increased

The magnitude of the electric field potential difference between the wingtips increases.

Explanation:

The magnitude of the electric potential difference is the induced emf and is given by the equation:

$emf = l (v \times B)$

where l = length

v = velocity

B = magnetic field

As the altitude of the airplane increases, the magnetic flux becomes stronger, the speed of the airplane becomes perpendicular to the magnetic field, i.e. $v \times B = vB sin90 = vB\\$ ,

the induced emf = vlB, and thus increases.

The magnitude of the electric field potential difference between the wingtips increases

3 0

3 years ago

3. A penguin waddles 8 m uphill before sliding back down to its friends in 2 seconds. If the penguin ends where it started, what

german

The velocity of penguin as he ends where he started was 0 m/s.

<h3>What is displacement?</h3>

Displacement is the length of straight line joining the initial and final position of the body.

Given is a penguin who waddled 8 m uphill before sliding back down to its friends in 2 seconds.

We know that the velocity is the rate of change of displacement with respect to time. Mathematically -

v = dx/dt

dx = v dt

∫dx = ∫v dt

Δx = vΔt

v = Δx/Δt

Now, the displacement of the penguin will be = Δx = 8 - 8 = 0

Then, its velocity will be -

v = 0/Δt = 0

Therefore, the velocity of penguin as he ends where he started was 0 m/s.

To solve more questions on kinematics, visit the link below-

brainly.com/question/27200847

#SPJ1

4 0

1 year ago