The initial speed at which the bomb would have to be ejected is approximately 36,260 m/s. The time of flight would be approximately 7,400 seconds.
To calculate the initial speed and time of flight of the volcanic bomb, we can use the equations of projectile motion. Let's break down the problem into its components:
Given:
Angle of projection (Co) = 40°
Vertical distance (h) = 3.70 km
Horizontal distance (d) = 8.30 km
Step 1: Calculate the initial vertical velocity (Vy0)
The vertical motion of the bomb is affected by gravity. We can use the equation:
Vy = Vy0 - gt
Where:
Vy is the vertical component of velocity
Vy0 is the initial vertical velocity
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time of flight
At the highest point of the bomb's trajectory, the vertical component of velocity becomes zero. Therefore, we can rewrite the equation as:
0 = Vy0 - gt
Solving for Vy0:
Vy0 = gt
Substituting the values:
Vy0 = (9.8 m/s^2)(3.70 km)
Convert 3.70 km to meters:
Vy0 = (9.8 m/s^2)(3,700 m)
Vy0 = 36,260 m/s
Step 2: Calculate the initial horizontal velocity (Vx0)
The horizontal motion of the bomb remains constant throughout its trajectory. We can use the equation:
Vx = Vx0
Where:
Vx is the horizontal component of velocity
Vx0 is the initial horizontal velocity
Since there is no acceleration in the horizontal direction, the initial horizontal velocity remains constant. Therefore, we can write:
Vx0 = Vx
Step 3: Calculate the initial speed (V0)
The initial speed of the bomb can be calculated using the Pythagorean theorem:
V0 = sqrt(Vx0^2 + Vy0^2)
Substituting the values:
V0 = sqrt((Vx)^2 + (Vy0)^2)
V0 = sqrt((Vx0)^2 + (gt)^2)
Since Vx0 = Vx, we can simplify the equation to:
V0 = sqrt((Vx)^2 + (gt)^2)
Step 4: Calculate the time of flight (t)
The time of flight can be calculated using the equation:
t = 2 * (Vy0 / g)
Substituting the values:
t = 2 * (36,260 m/s) / (9.8 m/s^2)
t = 7,400 s