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!

Where did the 4.9 come from??

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

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

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!!

To answer this question, we can use the concepts of projectile motion. We will break down the problem into different components and use equations of motion to find the required initial speed, time of flight, and the effect of air.

(a) To find the initial speed, we need to analyze the vertical and horizontal motion separately.

Vertical Motion:
We can use the equation for vertical displacement:
Δy = v₀y * t - (1/2) * g * t²

In this case, Δy (vertical displacement) is equal to -h (negative because the displacement is downwards), v₀y (vertical component of initial velocity) is v₀ * sin(θ), g is the acceleration due to gravity (-9.8 m/s²), and t is the time of flight.

Therefore,
-h = v₀ * sin(θ) * t - (1/2) * (-9.8) * t²
3.30 km = v₀ * sin(35°) * t + 4.9 * t²
Convert km to meters: 3.30 km = 3,300 m
3,300 m = v₀ * sin(35°) * t + 4.9 * t²

Horizontal Motion:
We can use the equation for horizontal displacement:
Δx = v₀x * t

In this case, Δx (horizontal displacement) is equal to d, v₀x (horizontal component of initial velocity) is v₀ * cos(θ), and t is the time of flight.

Therefore,
d = v₀ * cos(θ) * t
9.40 km = v₀ * cos(35°) * t
Convert km to meters: 9.40 km = 9,400 m
9,400 m = v₀ * cos(35°) * t

Now, we have a system of equations to solve simultaneously:
3,300 = v₀ * sin(35°) * t + 4.9 * t²
9,400 = v₀ * cos(35°) * t

Using these equations, we can find the initial speed v₀ and the time of flight t.

(b) To find the time of flight, once we determine v₀, we can substitute it back into either the vertical or horizontal motion equation from part (a) to solve for t.

(c) After solving part (a), we can discuss the effect of air. Ignoring air resistance, the calculations are based on ideal projectile motion. However, in reality, air resistance affects the trajectory of a projectile by reducing its height and range. Thus, the effect of air would decrease the values obtained in part (a) for the initial speed and time of flight.

To find the exact values and solve the system of equations, you can use numerical methods, such as substitution or elimination, or use a scientific calculator or computer software capable of solving simultaneous equations.