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 $${ \theta }_{ 0 }=35$$ 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 $$d=9.40 km$$? Ignore, for the moment, the effects of air on the bombs travel. (b) What would be the time of flight? (c) Would the effect of the air increase or decrease your answer in (a)?

Answer 1

(a) For projectile motion $$v_o$$ is given by formula

$${ v }_{ 0 }=\dfrac { x }{ cos{ \theta }_{ 0 } } \sqrt { \dfrac { g }{ 2(xtan{ \theta }_{ 0 }-y) } } $$

putting $$x = 9400 m$$, $$y = –3300 m$$, and $${ \theta }_{ 0 } = 35°$$.

$${ v }_{ 0 }=\dfrac { 9400 }{ \cos35^o } \sqrt { \dfrac { g }{ 2(x\tan35^o-3300) } } $$

$${ v }_{ 0 }= 255.5 ≈ 2.6 × { 10 }^{ 2 }m/s$$

(b) For projectile motion $$t$$ is given by formula
$$t=\dfrac { x }{ { v }_{ 0 }{ cos\theta }_{ 0 } } $$

$$t=\dfrac { 9400m }{ (255.5m/s)cos{ 35 }^{ o } } =45s.$$

(c) We expect the air to provide resistance but no appreciable lift to the rock, so we would need a greater launching speed to reach the same target.