a bomb is dropped from an airplane travelling horizontally with a speed of 300 mi/hr if the airplane is 10000ft. above the ground, how far from the target must it be released, neglect friction?
d = Vo*t + 0.5g*t^2 = 10.000 Ft.
0 + 4.9t^2 = 10,000,
t^2 = 2041,
t = Tf = 45.2 s. = Fall time = Time in
flight.
Dx = Xo * Tf = 300 m/s * 45.2 s. = 13,553 Ft.= Dist. it should be released from target.
Correction:
d = Vo*t + 0.5g*t^2 = 10,000 Ft.
0 + 16t^2 = 10,000,
t^2 = 625,
t = 25 s.
X0=300 mi/h * 1h/3600 s. = .0833 mi/s.
Dx = 0.08333 mi/s * 25s = 2.083 mi =
3333 m.
Where did you get the 16?
PURSIGIDO
Well, that's a blast of a question! Let's calculate it step by step. We know the initial velocity of the bomb horizontally is 300 mi/hr, and the height of the plane is 10,000 ft.
First, we need to convert the velocity to feet per second, because why not mix units like a DJ at a party? There are 5,280 feet in a mile and 60 minutes in an hour, so let's do some conversion magic!
300 mi/hr * 5280 ft/mi * 1 hr/3600 s = 440 ft/s
Now, let's look at how long it takes for the bomb to fall to the ground. We will use the free-fall equation:
h = (1/2) * g * t^2
Since we're neglecting friction, the only force acting on the bomb is gravity, so the acceleration (g) is approximately 32.2 ft/s^2.
10,000 ft = (1/2) * 32.2 ft/s^2 * t^2
Solving for t^2, we get:
t^2 = (2 * 10,000 ft) / 32.2 ft/s^2 = 621.12 s^2
Since we're only interested in the time taken to fall, we'll take the positive square root.
t ≈ √621.12 s ≈ 24.9 s
Now, we can calculate the horizontal distance traveled by the bomb during this time:
distance = velocity * time
distance ≈ 440 ft/s * 24.9 s ≈ 10,956 ft
So, releasing the bomb approximately 10,956 feet away from the target should make quite an explosive entrance! Just make sure to aim it properly.
To determine how far from the target the bomb must be released, we can use the principles of projectile motion. Assuming the target is on the same level as the ground, we can analyze the horizontal and vertical motion of the bomb separately.
First, let's consider the horizontal motion. Since the airplane is traveling horizontally with a speed of 300 mi/hr, the bomb will also have this horizontal velocity when released. The horizontal velocity remains constant throughout the motion, so we can disregard it when determining the distance.
Next, let's focus on the vertical motion. The bomb is dropped from a height of 10,000 ft above the ground. To find the time it takes for the bomb to reach the ground, we can use the equation:
h = (1/2) * g * t^2
Where:
h is the initial height (10,000 ft)
g is the acceleration due to gravity (approximately 32.2 ft/s^2)
t is the time it takes for the bomb to reach the ground.
Rearranging the equation, we get:
t^2 = (2h) / g
t^2 = (2 * 10,000) / 32.2
t^2 = 620.55
t ≈ 24.9 seconds
Now, we can find the horizontal distance by multiplying the time by the horizontal velocity:
distance = time * velocity
distance = 24.9 s * (300 mi/hr * 5280 ft/mi) / (3600 s/hr)
distance ≈ 7,311 ft
Therefore, neglecting friction, the bomb must be released approximately 7,311 feet horizontally from the target.