20 lbs of gun powder in a 14in. gun bullet travels 140000m at its maximum range

Question

20 lbs of gun powder in a 14in. gun bullet travels 140000m at its maximum range at 43 degrees. How long does it take the cannon ball to hit the ship? using t= 140000m/(Vo)(cos43) Yf=Yo + Voy + t +(1/2) ayt2 Yf= final velocity = 0 Voy= original velocity= Vosin43 Yo= initial position ay = acceleration = -9.8m/s2 20 lbs of gun powder in a 14in. gun bullet travels 140000m at its maximum range at 43 degrees. How long does it take the cannon ball to hit the ship? using t= 140000m/(Vo)(cos43) Yf=Yo + Voy + t +(1/2) ayt2 Yf= final velocity = 0 Voy= original velocity= Vosin43 Yo= initial position ay = acceleration = -9.8m/s2 Yf= final velocity = 0 Voy= original velocity= Vosin43 Yo= initial position ay = acceleration = -9.8m/s2

Explanation / Answer

Hi,

Given the cannon ball has its maximum range at 43o . We know that Range R = (vo2 / g) * Sin2      (where vo is the initial velocity)       => vo = (Rg/Sin2) = (140000 * 9.8 / 0.99756)                                    = 1172.76 m/s. Now the initial horizontal velocity = vxo = vo * Cos = 1172.76 * 0.73135                                                                               = 864.735 m/s. Hence the total time of travel = t = R/vxo = 140000/864.735 = 161.89 sec. Hope this helps you. Hope this helps you.

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