A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the spider is 0.880 m/s at an angle of 34.1° above the table, and it lands on the magazine 0.0790 s after leaving the table. Ignore air resistance. How thick is the magazine? Express your answer in millimeters.
To find the thickness of the magazine, we can use the vertical motion of the spider. Let's break down the problem and use the following equations of motion:
1. Vertical displacement (Δy) = (initial vertical velocity * time) + (0.5 * acceleration * time^2)
2. Initial vertical velocity (Vy) = initial velocity * sin(angle)
3. Acceleration (a) = -9.8 m/s^2 (due to gravity)
First, let's calculate the vertical displacement of the spider. We know that the initial vertical velocity (Vy) is given by:
Vy = initial velocity * sin(angle) = 0.880 m/s * sin(34.1°) = 0.456 m/s
The time (t) is given as 0.0790 s.
Using equation 1, we can rearrange it to solve for displacement:
Δy = (Vy * t) + (0.5 * a * t^2)
Δy = (0.456 m/s * 0.0790 s) + (0.5 * -9.8 m/s^2 * (0.0790 s)^2)
Δy = 0.036 m - 0.0308 m
Δy = 0.0052 m
Now, to convert the displacement from meters to millimeters, we multiply it by 1000:
Displacement in millimeters = 0.0052 m * 1000 = 5.2 mm
Therefore, the thickness of the magazine is 5.2 mm.