How, if at all, would the equations written in Parts C and E change if the projectile was thrown from the cliff at an angle above the horizontal? Explain your answer.

Answers

Answer 1
Answer:

Answer:

x = v₀ cos θ   t,   y = y₀ + v₀ sin θ t - ½ g t2

Explanation:

This is a projectile launch exercise, in this case we will write the equations for the x and y axes

Let's use trigonometry to find the components of the initial velocity

              sin θ = v_(oy) / v₀

              cos θ = v₀ₓ / v₀

              v_{y} = v_{oy} sin θ

              v₀ₓ = vo cos θ

now let's write the equations of motion

X axis

         x = v₀ₓ t

         x = v₀ cos θ   t

        vₓ = v₀ cos θ

Y axis

        y = y₀ + v_(oy) t - ½ g t2

        y = y₀ + v₀ sin θ t - ½ g t2

        v_{y} = v₀ - g t

       v_{y} = v₀  sin θ - gt

        v_(y)^(2) = v_{oy}^2 sin² θ - 2 g y

As we can see the fundamental change is that between the horizontal launch and the inclined launch, the velocity has components

Answer 2
Answer:

Final answer:

If a projectile is thrown at an angle above the horizontal, this introduces an additional vertical velocity component. This would affect the equations of Parts C and E by changing the values substituted into them, thereby altering the results for time of flight, highest point, and distance traveled.

Explanation:

If the projectile is thrown from a cliff at an angle above the horizontal, it would introduce a vertical component to the initial velocity in the equations. The equations of motion in Parts C and E would therefore need to incorporate this changed initial velocity.

In the equations, the initial vertical velocity is usually denoted by Vi*sin(θ), where θ is the angle at which the projectile is launched. Similarly, the horizontal velocity is expressed as Vi*cos(θ).

This additional vertical component would affect the time of flight, the height reached by the projectile, and the horizontal distance traveled before it hits the ground. Thus, while the form of the equations might not change, the values substituted into them would.

Learn more about Projectile Motion here:

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Which term describes the time it takes for an object to complete one full cycle of motion on a spring?

Answers

The term period (symbol: T) describes the time it takes for an object to complete one full cycle of motion on a spring.

The formula for time is: T  = 1 / f , where f is the frequency  , f= c / λ = wave speed c (m/s) / wavelength λ (m)..

The formula describes that as the frequency of a wave increases, the time period of the wave decreases.

There is no correct choice on the list you provided.

That time is called the "period" of the motion

Two balls, each with a mass of 0.844 kg,exert a gravitational force of 8.45 × 10−11 Non each other.How far apart are the balls? The value of the universal gravitational constant is 6.673 × 10−11 N m2/kg2.

Answers

The Universal Gravitation law is F=GMm/d^2
So: 8.45x10^-11=6,673x10^-11x0,844x0,844/d^2
d^2=1,78, so finale d=1,33m

A total solar eclipse is visible froma. all over Earth.
b. only within the moon’s umbra.
c. only within the moon’s penumbra.
d. only the dark side of Earth.

Answers

Answer:

b. only within the moon’s umbra.

Explanation:

Solar eclipse occurs when Moon comes between the Sun and the Earth in straight line and blocks the sunlight. A shadow of moon castes on the surface of the Earth. The region passing through the moon's umbra would view a total solar eclipse. The moon completely hides the surface of the sun. It would be visible to the people residing in the bright side of the Earth within the Umbra of the moon.

Only within the umbra of the moon's shadow.

what is the force of gravity between a 1000000000 kg asteroid and the earth when the asteroid is 3000000 m away from the earth?

Answers

Mass of the Earth = 5.972 x 10²⁴ kg
Mass of the asteroid = 1 x 10⁹ kg
Separation distance = 3 x 10⁶ m  (between the Earth's center and the asteroid's center)
                   

Gravitational force between two masses =

        (6.67 x 10⁻¹¹ nt-m²/kg²) · m₁ · m₂ / (distance)²

=  (6.67 x 10⁻¹¹ nt-m²/kg²) · (5.972 x 10²⁴ kg) ·  (1 x 10⁹ kg) / (3 x 10⁶ m)²

=  (6.67 · 5.972 / 9) x 10¹⁰ · (nt · m² · kg · kg / kg² · m²)

             =      13.28 x 10¹⁰     newtons

             =        132.8  Giga-newtons .

    
(about  14,926,763 tons)
         

PLEASE IS URGENT! "An object is decelerating. Suggest the resultant force acting on the object".

Answers

The resultant or the net force act on the object cannot be 0. it must have a value but the direction is opposite from the object move direction.

if the object move to the left, so the net force or the resultant is doing to the right

Please help hurry. It’s time.

Answers

Answer: a

Explanation: Hope it helps

Answer:

A

Explanation:

I think its a because A would be heavier making it go down faster than B. its like when you drop a father compared to a ball, the feather is lighter so it will go down slower than the ball because it has more mass