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Heads up, question today is incorrect (At least as of 2 AM). The reasoning below is correct until they solve d = .5at^2. For d = 2000 m, t = 20.. they say 40. The correct answer should be 400 m instead of 800 m.
"Question:A projectile is fired from a cannon with a velocity of 400 m/s at an angle of 30° to the horizontal. If the cannon is instead moving at a horizontal velocity of 10 m/s (in the same direction as the projectile) when the projectile launches, how much farther will the projectile travel? Neglect air resistance.
Explanation: The first step is to identify how long the projectile will spend in the air. This will be based off of its vertical velocity 400*sin(°30) = 200 m/s. Using vf2 = vi2 + 2ad, we get d = 2000 m. This represents the peak height reached by the projectile. The time to reach the peak would then be calculated with d = (1/2)at2, which equals 40 s. However, this is the time to reach the peak; the time to fall back down would be another 40 s. Thus the projectile spends 80 sec in flight.
We do not need to calculate how far the projectile will travel in this time, only how much farther it will travel if it has an extra 10 m/s to its horizontal velocity. This extra 10 m/s adds to the distance traveled linearly since d = vt. Therefore we can simply say d = (10)(80) = 800 m."
"Question:A projectile is fired from a cannon with a velocity of 400 m/s at an angle of 30° to the horizontal. If the cannon is instead moving at a horizontal velocity of 10 m/s (in the same direction as the projectile) when the projectile launches, how much farther will the projectile travel? Neglect air resistance.
Explanation: The first step is to identify how long the projectile will spend in the air. This will be based off of its vertical velocity 400*sin(°30) = 200 m/s. Using vf2 = vi2 + 2ad, we get d = 2000 m. This represents the peak height reached by the projectile. The time to reach the peak would then be calculated with d = (1/2)at2, which equals 40 s. However, this is the time to reach the peak; the time to fall back down would be another 40 s. Thus the projectile spends 80 sec in flight.
We do not need to calculate how far the projectile will travel in this time, only how much farther it will travel if it has an extra 10 m/s to its horizontal velocity. This extra 10 m/s adds to the distance traveled linearly since d = vt. Therefore we can simply say d = (10)(80) = 800 m."
