Question

During volcanic eruptions, chunks of solid rock can be blasted out of the volcano; these projectiles are called volcanic bombs. (a) At what initial speed would a bomb have to be ejected, at angle 35degrees to the horizontal, from the vent at A in order to fall at the foot of the volcano at B, at vertical distance h=3.30km and horizontal distance d=9.40km? Ignore, for the moment, the effects of air on the bomb's travel. (b) What would be the time of flight? (c) Would the effect of the air increase or decrease in your answer in (a)?

Thanks in advance!

Answer 1

fall 3300 m

range = 9400 m

speed = s

u = constant horizontal speed = s cos 35
= .819 s
if t is time in air then
t = 9400 / .819 s = 11,474/s

Vi = initial vertical speed = s sin 35 = .574 s

z = altitude

z = 3300 + Vi t - 4.9 t^2
ground is z = 0
so if t is the time in the air then
4.9 t^2 - Vi t - 3300 = 0

4.9(11,474/s)^2 - .574 s(11,474/s) -3300 = 0

4.9(11,474/s)^2 = 9886
11,474/s = 44.9
s = 255 meters/second

t = 11,474/s = 44.9 seconds

check my algebra !

if you have air drag, you need higher muzzle velocity

Answer 2

9886, where does it come from? How did you get it?

Answer 3

Where did the 4.9 come from??

Answer 4

During volcanic erup-
tions, chunks of solid rock
can be blasted out of the vol-
cano; 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 u0 􏰁 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
Please help me!!