Question

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 60 m away, making a 30° angle with the ground. How fast was the arrow shot?

Accepted Answer

Step 1: Break the problem into horizontal and vertical motion components. The arrow's motion can be analyzed as a projectile motion problem. The horizontal distance traveled by the arrow is 60 m, and the angle it makes with the ground upon impact is 30°. Assume the arrow was shot parallel to the ground, so the initial vertical velocity is 0 m/s. Step 2: Use the angle of impact to determine the relationship between the horizontal and vertical velocities at the moment the arrow hits the ground. The tangent of the angle of impact is given by $$ 	an(	heta) = \frac{v_y}{v_x} $$, where $$ v_y  is $$the vertical velocity and $$ v_x  is $$the horizontal velocity. Here, $$ 	heta = 30° . $$Step 3: Relate the vertical velocity $$ v_y  to $$the time of flight using the kinematic equation for vertical motion: $$ v_y = g \cdot t $$, where $$ g  is $$the acceleration due to gravity (9.8 m/s²) and $$ t  is $$the time of flight. Combine this with the relationship from Step 2 to express $$ t  in $$terms of $$ v_x . $$Step 4: Use the horizontal motion equation to find the time of flight. The horizontal distance traveled is given by $$ x = v_x \cdot t $$, where $$ x = 60 \; 	ext{m} . $$Solve for $$ t  in $$terms of $$ v_x . $$Step 5: Combine the equations from Steps 3 and 4 to solve for $$ v_x $$, the horizontal velocity, which is also the initial velocity of the arrow since it was shot parallel to the ground. Substitute known values ($$ g = 9.8 \; 	ext{m/s}^2 $$, $$ 	an(30°) $$, and $$ x = 60 \; 	ext{m} ) to $$find the final expression for the initial velocity.

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Key Concepts

Projectile Motion

Kinematic Equations

Trigonometry in Physics