A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the spider is 0.840 m/s at an angle of 36.0° above the table, and it lands on the magazine 0.0710 s after leaving the table. Ignore air resistance. How thick is the magazine? Express your answer in millimeters.
so far i have 0.010354612
NVM the answer was 10.35 i just had to express in mm
To find the thickness of the magazine, we can use the vertical motion of the spider when it jumps onto the magazine. We'll need to calculate the vertical displacement during its flight time.
1. Start by analyzing the vertical component of the spider's initial velocity.
- Given: Initial velocity (Vi) = 0.840 m/s at an angle of 36.0° above the table
- Vertical component of the initial velocity (ViY) = Vi * sin(θ)
ViY = 0.840 m/s * sin(36.0°)
2. Calculate the vertical displacement of the spider during its flight time.
- Given: Time of flight (t) = 0.0710 s
- Displacement (dY) = ViY * t + (1/2) * g * t^2
(where g is the acceleration due to gravity, approximately 9.81 m/s^2)
dY = ViY * t + (1/2) * 9.81 m/s^2 * (t^2)
3. Once we have the vertical displacement, we can find the thickness of the magazine.
- The thickness of the magazine is equal to the vertical displacement (dY).
- Convert the displacement to millimeters (mm).
Now, let's calculate the thickness of the magazine using the given values.
1. Calculate ViY:
ViY = 0.840 m/s * sin(36.0°)
2. Calculate dY:
dY = ViY * t + (1/2) * 9.81 m/s^2 * (t^2)
3. Calculate the thickness:
Convert dY to millimeters by multiplying it by 1000.
Please provide the values you have for Vi, θ (angle), t, ViY, dY, and your calculations so far, and I can help you complete the problem.