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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