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. Figure 4-51 shows a cross section of Mt. Fuji, in Japan. (a) At what initial speed would a bomb have to be ejected, at angle ?0 = 38° to the horizontal, from the vent at A in order to fall at the foot of the volcano at B, at vertical distance h = 3.80 km and horizontal distance d = 10.9 km? Ignore, for the moment, the effects of air on the bomb's travel. (b) What would be the time of flight?

Explanation / Answer

write the y(t) equation of motion for the bomb y(t)=y0+v0y t - 1/2 gt^2 y0=initial height =2800m v0y = initial vertical velocity = v0 sin 44 g=9.8m/s/s t=time we also know that if the object travels a horizontal distance of 8800m, and we are ignoring air friction, we can write horizontal dist = vhor x time since there are no forces acting in the horizontal direction which means the hor vel is constant this allows us to write time as t=hor dist/v hor = 8800m/v0 cos 44 substitute this into the y(t) equation: y(t)=2800 + v0 sin 44(8800/cos 44) - 1/2 g(8800/v0 cos 44)^2 you are interested in the case that y=0, so set y=0 and solve for v0 once you have v0, you can find time from 8800m = v0 cos 44 t

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.