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 shows a cross section of Mt. Fuji, in Japan. (a) At what initial speed would a bomb have to be ejected, at angle ?0 = 33? to the horizontal, from the vent at A in order to fall at the foot of the volcano at B, at vertical distance h = 3.30 km and horizontal distance d = 8.10 km? Ignore, for the moment, the effects of air on the bomb's travel. (b) What would be the time of flight?

Explanation / Answer

Your answer can be ….. • u is the vertical velocity and v is the horizontal velocity. Time to travel 9.2 km is t = 8.10/v. Time and displacement of -3.3m is given by' -3.3 = ut - 0.5*9.8 t^2 solving for t -3.8 = 8.10u /v- 0.5*9.8 (8.10/v.)^2 u/v = tan 33, Hence -3.8 = (8.10*tan 33) - (0.5*9.8)* (8.10/v) ^2 now for v v = ±6.5m/s. Since it was going up initially v = 6.5m/s u = v*tan 33 = 6.5* tan 33 = 4.22m/s initial speed.. Resultant of u and v is v (6.5^2 + 4.22^2) = 7.8 m/s b) time of flight Time t = 8.10/v. = 8.10/6.5 = 1.246 s

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