An archer shoots an arrow with an initial speed of 100m/s at an angle of 30^circ from the horizontal. How high does the arrow travel? 510 m 380 m 130 m ) 0.2 m

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An archer shoots an arrow with an initial speed of 100m/s at an angle of 30^circ from the horizontal. How high does the arrow travel? 510 m 380 m 130 m ) 0.2 m

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## Step 1: Identify the components of the initial velocity ### The initial speed is $100 \, \text{m/s}$ at an angle of $30^\circ$. We need to find the vertical component of the initial velocity using $v_{0y} = v_0 \sin(\theta)$. ## Step 2: Calculate the vertical component of the initial velocity ### Using $v_{0y} = 100 \sin(30^\circ) = 100 \times 0.5 = 50 \, \text{m/s}$. ## Step 3: Use the kinematic equation to find the maximum height ### The kinematic equation for vertical motion is $v_y^2 = v_{0y}^2 - 2g h$, where $v_y = 0$ at the maximum height. Solving for $h$, we get $h = \frac{v_{0y}^2}{2g}$. ## Step 4: Substitute the known values into the equation ### $h = \frac{(50 \, \text{m/s})^2}{2 \times 9.8 \, \text{m/s}^2} = \frac{2500 \, \text{m}^2/\text{s}^2}{19.6 \, \text{m/s}^2} = 127.55 \, \text{m}$. ## Step 5: Round to the nearest option ### The closest option to $127.55 \, \text{m}$ is $130 \, \text{m}$.

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An archer shoots an arrow with an initial speed of 100m/s at an angle of 30^circ from the horizontal. How high does the arrow travel? 510 m 380 m 130 m ) 0.2 m

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