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 in order to make a direct hit? neglect the resistance?

16t^2 = 10,000.

t = 25 s.
d = 300mi/h * (25/3600)h =

To solve this problem, we need to consider the horizontal motion and vertical motion of the bomb.

First, let's focus on the horizontal motion. Since the airplane is traveling horizontally with a speed of 300 mi/hr, the bomb will have the same horizontal velocity. We can convert this velocity to feet per second to match the unit system of the altitude.

1 mile = 5280 feet
1 hour = 3600 seconds

So, the horizontal velocity is:

300 mi/hr * (5280 ft/mi) / (3600 s/hr) = 440 ft/s

Now, let's consider the vertical motion of the bomb. The distance the bomb falls vertically can be calculated using the equation:

h = (1/2) * g * t^2

Where:
h is the vertical distance (10,000 ft),
g is the acceleration due to gravity (32.2 ft/s^2), and
t is the time it takes for the bomb to fall.

Now, we can solve for t:

10,000 ft = (1/2) * (32.2 ft/s^2) * t^2

Dividing both sides by (1/2) * (32.2 ft/s^2), we get:

t^2 = (10,000 ft) / (0.5 * 32.2 ft/s^2)

t^2 = 621.12 s^2

Taking the square root of both sides:

t ≈ 24.92 s

Since the bomb is dropped from the airplane, both the horizontal and vertical motions occur for the same duration of time. Therefore, the horizontal distance covered by the bomb can be calculated using the equation:

distance = horizontal velocity * time

distance = 440 ft/s * 24.92 s ≈ 10959.68 ft

Therefore, the bomb must be released approximately 10959.68 ft (or about 2.08 miles) from the target to make a direct hit, neglecting air resistance.

how long does it take to fall 10000 ft?

16t^2 = 10000
t = 25
How far will the plane travel in 25 seconds?