a plane moving horizontally at 120 m/s drops a package from an elevation of 240 m. How far did the package go before it hit the ground?

Step by step directions please!

thank you!!

Step 1: Identify the given information:

- Initial horizontal velocity of the plane (Vx) = 120 m/s
- Initial vertical position (height) of the package (h) = 240 m

Step 2: Determine the time it takes for the package to hit the ground using the equations of motion. For vertical motion, we can use the equation: h = Vyi * t + (1/2) * a * t^2, where:
- Vyi is the initial vertical velocity (which is zero because the package is dropped)
- a is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time it takes for the package to hit the ground

Since the initial vertical velocity is zero, the equation simplifies to:
h = (1/2) * a * t^2

Plugging in the given value of h = 240 m and a = 9.8 m/s^2:
240 = (1/2) * 9.8 * t^2

Solving for t^2:
480 = 9.8 * t^2
t^2 = 480 / 9.8
t^2 = 48.98
t ≈ √48.98
t ≈ 7 seconds (rounded to 2 decimal places)

Step 3: Calculate the horizontal distance traveled by the package during the time it takes to hit the ground. The horizontal distance (d) can be determined using the equation: d = Vx * t, where Vx is the initial horizontal velocity of the plane and t is the time calculated in Step 2.

Plugging in the given values of Vx = 120 m/s and t = 7 seconds:
d = 120 * 7
d = 840 meters

Therefore, the package traveled approximately 840 meters horizontally before hitting the ground.

Sure! To find out how far the package went before it hit the ground, we can use the equation of motion for objects in free fall. Here are the step-by-step directions:

Step 1: Recall the equation of motion for objects in free fall:
- s = ut + (1/2)gt^2
- Here, s represents the vertical displacement (distance), u is the initial vertical velocity, t is the time taken, and g is the acceleration due to gravity.

Step 2: Identify the given information:
- The plane is moving horizontally at 120 m/s.
- The package is dropped from an elevation of 240 m.
- Since the plane is moving horizontally, there is no initial vertical velocity, u = 0 m/s.
- The acceleration due to gravity is approximately 9.8 m/s^2 (usually taken as -9.8 m/s^2 for upward direction).

Step 3: Find the time taken for the package to reach the ground:
- Since the package is dropped vertically, the initial vertical velocity is zero, u = 0 m/s.
- The final vertical displacement is the distance it falls, s = -240 m (taking downward as negative).
- Use the equation of motion s = ut + (1/2)gt^2, we can rearrange it to solve for t:
- t = √((2s)/g)

Step 4: Calculate the time taken (t):
- Substitute s = -240 m and g = -9.8 m/s^2 into the equation:
- t = √((2 * (-240)) / (-9.8))

Step 5: Evaluate the time taken:
- Use a calculator to solve the equation:
- t ≈ 7.93 s (rounded to two decimal places)

Step 6: Find the horizontal distance traveled by the package:
- Since the plane is moving horizontally at a constant speed, the horizontal distance is given by the equation:
- distance = speed * time

Step 7: Calculate the horizontal distance:
- Substitute the given speed of the plane, 120 m/s, and the calculated time, 7.93 s, into the equation:
- distance = 120 m/s * 7.93 s

Step 8: Evaluate the horizontal distance:
- Use a calculator to solve the equation:
- distance ≈ 951.6 m (rounded to one decimal place)

Therefore, the package went approximately 951.6 meters horizontally before hitting the ground.

How long was it falling?

It fell 240 m with no initial vertical velocity
a = -g = -9.8 m/s^2
Vo = 0
Zo = 240 m
Zfinal = 0 m
so
0 = 240 + 0 - (1/2)(9.8) t^2
4.9 t^2 = 240
t = 7 seconds in the air
NOW
there are no horizontal forces on it so it keeps going ahead at 120 m/s for the 7 seconds, moving 840 m horizontal.

By the way unless it turned or changed speed, the plane also moved 840 m during the 7 seconds and the bomb burst right under the plane. It is best to turn and ascend after dropping a bomb.