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3 years ago

For a given launch velocity, at which launch angle does the projectile undergo the maximum horizontal displacement

Physics

2 answers:

aliya0001 [1]3 years ago

8 0

Answer:

45°

Explanation:

The horizontal displacement of any projectile (assuming the it is launched from the same height as it arrives) is given by:

$X = \frac{2*V^2}{g}*cos\alpha *sin\alpha$

As you can see it depends on both the sine and cosine of the launch angle.

When you graph that multiplication (cos(α) * sin(α)), you can see that it has a maximum peak at $\alpha =45\°$ (shown in the picture)

You could also derive the expression and find its maximum but I'll leave that for another time.

-Dominant- [34]3 years ago

4 0

C 45 is the answer :)

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A 700-kg car, driving at 29 m/s, hits a brick wall and rebounds with a speed of 4.5 m/s. what is the car's change in momentum du

Viefleur [7K]

The change in the momentum of the car due to collision between car and wall is $\fbox{\begin\ -23450\text{ kg.m}/\text{s}\end{minispace}}$ or $\fbox{\begin\\-2.3450 \times {10^4}\,\text{kg.m}/\text{s}\end{minispace}}$ .

Further Explanation:

Let us consider the car is moving towards the right direction and it collides with the wall. After the collision, the car will bounce back or car will rebound opposite to the direction of its motion i.e., towards left. Therefore, the final velocity of the car is opposite to the direction of the initial velocity of the car.

Given:

The mass of the car is $700\,{\text{kg}}$ .

The velocity of the car before collision is $29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

The velocity of the car after collision is $4.5\text{ m}/\text{s}$ .

Concept:

The momentum of an object is defined as the product of mass of object and the velocity with which the object is moving.

The initial momentum of the car is:

$\fbox{\begin\\{p_i}= m{v_i}\end{minispace}}$ …… (I)

Here, ${p_i}$ is the initial momentum of the car, $m$ is the mass of the car and ${v_i}$ is the initial velocity of the car.

Substitute $700\,{\text{kg}}$ for $m$ and $29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${v_i}$ in equation (I).

$\begin{gathered}{p_i} = \left( {700\,{\text{kg}}} \right) \cdot \left( {29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}} \right) \\ = 20300\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\ \end{gathered}$

The final momentum of the car is defined as the product of mass of car and the velocity of the car after collision or final velocity of the car.

The final momentum of the car is:

$\fbox{\begin\\{p_f} = m{v_f}\end{minispace}}$ …… (II)

Here, ${p_f}$ is the final momentum and ${v_f}$ is the final velocity.

Substitute $700\,{\text{kg}}$ for $m$ and - $4.5\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${v_f}$ in equation (II).

$\begin{aligned}{p_f}&=\left( {700\,{\text{kg}}} \right)\cdot\left( {- 4.5\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}}{\text{s}}}} \right.\kern-\nulldelimiterspace}{\text{s}}}}\right)\\&=-3150\,{{{\text{kg}} \cdot {\text{m}}}\mathord{\left/{\vphantom{{{\text{kg}}\cdot {\text{m}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

The change in the momentum of the car after collision is the difference between the momentum of car before collision and the momentum of car after collision.

The change in momentum of the car is:

$\fbox{\begin\Delta p = {p_f} - {p_i}\end{minispace}}$ …… (III)

Here, $\Delta p$ is the change in the momentum of the car.

Substitute $- 3150\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${p_f}$ and $20300\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${v_i}$ in equation (III).

$\begin{aligned}\Delta{p}&=-3150\text{ kg}.\text{m}/\text{s}\ -20300\text{ kg}.\text{m}/\text{s}\\&=-23450\text{ kg}.\text{m}/\text{s}\end{aligned}$

Thus, the change in the momentum of the car due to collision between car and wall is $\fbox{\begin\ -23450\text{ kg.m}/\text{s}\end{minispace}}$ or $\fbox{\begin\\-2.3450 \times {10^4}\,\text{kg.m}/{s}\end{minispace}}$ .

Learn more:

1. The motion of a body under friction brainly.com/question/4033012/

2. A ball falling under the acceleration due to gravity brainly.com/question/10934170/

3.Conservation of energy brainly.com/question/3943029/

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Change in momentum, collision, initial velocity, final velocity, initial momentum, final momentum,-23450 kgm/s^2, -23450 kgm/s2, -2.3450*10^6 kgm/s^2, -2.3450*10^6 kgm/s2.

7 0

3 years ago

Integrated Concepts A lightning bolt strikes a tree, moving 20.0 C of charge through a potential difference of 1.00×102 MV . (a)

vodka [1.7K]

Answer:

a) 2*10^9 J

b) 764.8 kg

Explanation:

Given that

Energy of charge, q = 20 C

Potential difference, ΔV = 1*10^2 MV = 1*10^2 * 10^6 V = 1*10^8 V

To find the energy dissipated, we use the formula

ΔU = qΔV

ΔU = 20 * 1*10^8

ΔU = 2*10^9 J

Change in temperature, ΔT = 100 - 15°

ΔT = 85° C

Change in energy, ΔU = 2*10^9 J

Specific heat of water, C = 4180 j./Kg.K

Latent heat of vaporization, L(v) = 2.26*10^6 J/Kg

Q1 = mcΔT

Q2 = mL(v)

Net energy needed, U = Q1 + Q2

U = mcΔT + mL(v)

U = m (cΔT + L(v))

m = U /[cΔT + L(v)]

Being that we have all the values, we then substitute

m = 2*10^9 / [(4180 * 85) + 2.26*10^6]

m = 2*10^9 / (3.553*10^5 + 2.26*10^6]

m = 2*10^9 / 2.615*10^6

m = 764.8 kg

Having 765 kg of steam at the temperature would have extreme effect on the tree, damaging it permanently. Possibly even blowing it to pieces

6 0

3 years ago

Help ASAP <br> Just answer the first question for me please!

Usimov [2.4K]

Answer:

Im pretty sure its b

4 0

3 years ago

Every chemical element goes through natural exponential decay, which means that over time its atoms fall apart. The speed of eac

Naddik [55]

Answer:

t = (ti)ln(Ai/At)/ln(2)

t = 14ln(16)/ln(2)

Solving for t

t = 14×4 = 56 seconds

Explanation:

Let Ai represent the initial amount and At represent the final amount of beryllium-11 remaining after time t

At = Ai/2^n ..... 1

Where n is the number of half-life that have passed.

n = t/half-life

Half life = 14

n = t/14

At = Ai/2^(t/14)

From equation 1.

2^n = Ai/At

Taking the natural logarithm of both sides;

nln(2) = ln(Ai/At)

n = ln(Ai/At)/ln(2)

Since n = t/14

t/14 = ln(Ai/At)/ln(2)

t = 14ln(Ai/At)/ln(2)

Ai = 800

At = 50

t = 14ln(800/50)/ln(2)

t = 14ln(16)/ln(2)

Solving for t

t = 14×4 = 56 seconds

Let half life = ti

t = (ti)ln(Ai/At)/ln(2)

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4 years ago

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Pavel [41]

Casts? When you get injured ?

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