An arrow is shot at 28.0° above the horizontal. Its initial speed is 48 m/s and it hits the target.?
The target is at the height from which the arrow was shot. How far away is it?
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answers:
Fabio _ say: Your parameters are incomplete. The aerodynamic drag of the arrow also needs to be considered.
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Tiburon say: very far
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oubaas say: range (distance) = V^2/g*sin 56° = 48^2/9.806*0.829 = 194.8 m
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electron1 say: Since the target is at the height from which the arrow was shot, you can use the following equation to determine the horizontal distance the arrow moved.
Range = v^2/g * sin 2θ
Range = 48^2/9.8 * sin 56
This is approximately 195 meters.
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RealPro say: The last time you asked this looking for maximum height, you were given the formula for maximum height.
This time you need the formula for range. Maybe you could try googling it yourself first?
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