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A physicist is testing a brand new projectile launcher in her lab that is designed to launch human skulls in a vacuum. She can adjust the angle to whatever she wants. She tests it by launching a skull at an angle of 70 degrees above the horizontal at a velocity of 18.0m/s at ground level.

A. What are the x and y components of the skull's initial velocity?

B. How fast is the skull traveling at its peak height?

C. How long until the skulls hits the floor?

D. What is the range of the skull?

User Nunners
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A physicist is trying out a new machine in her lab. It can launch human skulls in a space without air. She can change the way it points. She tries it by shooting a skull at the ground. She shoots it up at a 70-degree angle and it goes 18 meters per second.

A. What is the speed of the skull when it goes up and when it goes sideways?

B. How fast is the skull going when it is as high as it can go?

C. How much time passes before the skull hits the ground?

D. How far away does the skull go?

To answer the questions related to this physics problem, you can use the following formulas:

The x-component of the initial velocity (vx) can be found using the formula: vx = v0*cos(theta), where v0 is the initial velocity magnitude (18.0 m/s) and theta is the launch angle (70 degrees).

The y-component of the initial velocity (vy) can be found using the formula: vy = v0*sin(theta).

The time it takes for the skull to hit the ground (t) can be found using the formula: t = (2*vy)/g, where g is the acceleration due to gravity (9.81 m/s^2).

The maximum height reached by the skull (h) can be found using the formula: h = (vy^2)/(2*g).

The range of the skull (R) can be found using the formula: R = (vx*vy)/g.

A. To find the x and y components of the skull's initial velocity, we use the following equations:

Vx = V cosθ

Vy = V sinθ

where V is the magnitude of the velocity, θ is the launch angle, Vx is the velocity in the x-direction, and Vy is the velocity in the y-direction.

Given V = 18.0 m/s and θ = 70 degrees, we have:

Vx = 18.0 m/s × cos(70 degrees) ≈ 5.53 m/s

Vy = 18.0 m/s × sin(70 degrees) ≈ 17.5 m/s

Therefore, the x and y components of the skull's initial velocity are 5.53 m/s and 17.5 m/s, respectively.

B. At the peak height, the skull's vertical velocity (Vy) will be zero. We can use the following equation to find the speed at the peak height:

Vy = V sinθ - gt

where g is the acceleration due to gravity (approximately 9.81 m/s^2).

Setting Vy to zero and solving for V, we have:

V = gt/sinθ ≈ 36.5 m/s

Therefore, the skull is traveling at a speed of approximately 36.5 m/s at its peak height.

C. To find the time it takes for the skull to hit the floor, we can use the following equation:

y = Vyt - 0.5gt^2

where y is the initial height of the skull (assumed to be zero), Vy is the initial velocity in the y-direction, g is the acceleration due to gravity, and t is the time.

Setting y to zero and solving for t using the quadratic formula, we get:

t = [Vy + sqrt(Vy^2 + 2gy)]/g

Substituting the values we have, we get:

t = [17.5 m/s + sqrt((17.5 m/s)^2 + 2 × 9.81 m/s^2 × 0)]/9.81 m/s^2 ≈ 1.79 s

Therefore, the skull will hit the floor after approximately 1.79 seconds.

D. To find the range of the skull, we can use the following equation:

x = Vx × t

Substituting the values we have, we get:

x = 5.53 m/s × 1.79 s ≈ 9.89 m

Therefore, the range of the skull is approximately 9.89 meters.

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User Shonni
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A. To find the x and y components of the skull's initial velocity, we first need to break the initial velocity vector into its x and y components. The initial velocity vector can be represented as:

  • v₀ = 18.0 m/s at an angle of 70 degrees above the horizontal

The x component of the initial velocity can be found using:

  • v₀x = v₀cosθ

where θ is the angle above the horizontal. Substituting the values, we get:

  • v₀x = 18.0 m/s cos(70°) = 5.60 m/s

The y component of the initial velocity can be found using:

  • v₀y = v₀sinθ

Substituting the values, we get:

  • v₀y = 18.0 m/s sin(70°) = 17.22 m/s

Therefore, the x and y components of the skull's initial velocity are 5.60 m/s and 17.22 m/s, respectively.

B. At the peak height, the vertical component of the velocity becomes zero. Therefore, we can use the equation:

  • v² = v₀² - 2gy

where v₀ is the initial velocity, g is the acceleration due to gravity, and y is the vertical displacement. At the peak height, y is equal to:

  • y = (v₀y)² / (2g)
  • Substituting the values, we get:
  • y = (17.22 m/s)² / (2*9.81 m/s²) = 15.00 m

Using this value of y and the initial velocity, we can find the speed of the skull at the peak height:

  • v = √(v₀x² + 2gy)

Substituting the values, we get:

  • v = √((5.60 m/s)² + 2*(9.81 m/s²)*(15.00 m)) = 22.15 m/s

Therefore, the skull is traveling at a speed of 22.15 m/s at its peak height.

C. To find the time it takes for the skull to hit the floor, we can use the equation:

y = v₀yt - 1/2gt²

where t is the time taken for the skull to hit the floor. At the point of impact, y is equal to zero. Substituting the values, we get:

  • 0 = (17.22 m/s)t - 1/2*(9.81 m/s²)t²

Solving for t, we get:

  • t = 3.53 s

Therefore, it takes 3.53 seconds for the skull to hit the floor.

D. To find the range of the skull, we can use the equation:

  • x = v₀x * t

Substituting the values, we get:

  • x = (5.60 m/s) * (3.53 s) = 19.77 m

Therefore, the range of the skull is 19.77 meters.

User Lupin
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