During volcanic eruptions, chunks of solid rock can be blasted out of the volcan

Question

During volcanic eruptions, chunks of solid rock can be blasted out of the volcano; these projectiles are called volcanic bombs. The figure below shows a cross section of Mt Fuji in Japan, At what initial speed would a bomb have to be ejected, at 35 degree to the horizontal, from the vent at A in order to fall at the foot of the volcano at B? Ignore, for the moment, the effects of air on the bomb's travel, What would be the time of flight? Would the effect of the air increase or decrease your answer in (a)?

Explanation / Answer

let the velocity of rock be V a) Horizontal velocity = vcos35 horizontal distance = 9.4km time taken = T = 9400/vcos35 take upward direction to be +ve Vertical velocity = vsin35 vertical distance = -3.3km vertical acceleration = g= -9.8 Time of flight = T =9400/vcos35 now we have the equation s= ut +.5at^2 -3300 = vsin35*9400/vcos35 + .5(-9.8) *(9400/vcos35)^2 =>-3300 - 6581.95 = -645242178/v^2 =>v = square root (645242178/(3300 + 6581.95)) =255.52m/sec b) time of flight =T = 9400/Vcos35 =44.91sec c) effect of air will increase my answer as there will be a drag along both the direction decreasing both the velocities and hence increasing the required time and required initial velocity

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