A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the spider is 0.890 m/s at an angle of 36.0° above the table, and it lands on the magazine 0.0660 s after leaving the table. Ignore air resistance. How thick is the magazine? Express your answer in millimeters.
well, the vertical position (in meters) of the spider at time t is
y(t) = 0.890 sin 36° t - 4.9t^2
= 0.523t - 4.9t^2
So, what is y(0.0660)?
To find the thickness of the magazine, we can analyze the vertical motion of the spider. Since we ignore air resistance, the only force acting on the spider is gravity.
First, let's break down the initial velocity of the spider into vertical and horizontal components. We can use trigonometry to find the vertical component of the initial velocity:
Vertical component: v_y = v_initial * sin(angle)
v_y = 0.890 m/s * sin(36.0°)
v_y = 0.890 m/s * 0.5878
v_y = 0.522 m/s
Next, we can determine the time it takes for the spider to reach the magazine. We are given that the time is 0.0660 seconds.
Now, we can use the equation of motion for vertical motion:
Vertical distance: Δy = v_y * t + 0.5 * a * t^2
Since the spider starts from rest vertically (at the table) and lands on the magazine, its vertical distance is equal to the thickness of the magazine.
Since we are ignoring air resistance, the acceleration due to gravity, 'a', is approximately -9.8 m/s^2 (considering downward direction as negative).
So, the equation becomes:
Thickness of magazine: Δy = 0 + 0.5 * (-9.8 m/s^2) * (0.0660 s)^2
Δy = 0.5 * (-9.8 m/s^2) * (0.004356 s^2)
Δy = -0.0214 m
Note: The negative sign indicates the downward direction.
Finally, to express the thickness of the magazine in millimeters, we can convert the result to millimeters by multiplying by 1000:
Thickness of magazine in millimeters: -0.0214 m * 1000 mm/m
Thickness of magazine in millimeters: -21.4 mm
The thickness of the magazine is 21.4 millimeters.