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Projectile Motion: Calculating Initial Speed for Arrow at 30° Angle
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[QUOTE="Epic Mango, post: 4945886, member: 535515"] The trick to solving these problems are detaching horizontal movement and vertical movement. What I mean by this is thinking about each dimension separately. Okay, so we first list what we know. θ = 30 degrees, d = 100 meters, g = -9.8 meters per second squared We also need these kinematics equations: d = v[SUB]i[/SUB] * t + 1/2 * a * t^2 v = v[SUB]i[/SUB] + a*t We then think about the problem in the horizontal. There is no acceleration in the horizontal, so we use our first kinematics equation: d = v[SUB]h[/SUB]*t We can rewrite v[SUB]horizontal[/SUB] as v[SUB]original[/SUB] * cos(θ). Now, the equation for horizontal distance is: d = v[SUB]i[/SUB] * cos(θ) * t. Now, we have to determine t. We can't determine the t from the horizontal, so we switch over to thinking in the vertical. We use the second kinematics equation, using gravity as our acceleration. At the peak of the arrow's path, v[SUB]v[/SUB] = 0. The equation is now: v[SUB]i[/SUB] * sin(θ) - 9.8 * t[SUB]top[/SUB] = 0 However, the fact remains that, at the moment, we are solving for the time it takes to reach the top of the arrow's path. We need the time it takes to hit the ground. We know that it takes the same amount of time to go up as the amount of time to go down. So, the equation for t is: t = t[SUB]top[/SUB] * 2 t[SUB]top[/SUB] = t/2 Now, we plug that into v[SUB]i[/SUB] * sin(θ) - 9.8 * t[SUB]top[/SUB] = 0. Rewrite the equation to solve for t. Plug the equation for t we just found into the equation for horizontal distance: d = v[SUB]i[/SUB] * cos(θ) * t I trust you can take over from here. Hope this helped! [/QUOTE]
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Projectile Motion: Calculating Initial Speed for Arrow at 30° Angle
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