A plane is flying horizontally with speed

193 m/s at a height 2010 m above the ground, when a package is dropped from the plane.The acceleration of gravity is 9.8 m/s^2.

1.What is the horizontal distance from the release point to the impact point?
Answer in units of m.

2.A second package is thrown downward from the plane with a vertical speed v1 = 58 m/s.
What is the magnitude of the total velocity of the package at the moment it is thrown as seen by an observer on the ground?
Answer in units of m/s.

3.What horizontal distance is traveled by this package?
Answer in units of m

1. 0.5g*t^2 = 2010 m.

4.9t^2 = 2010.
t = 20.3 s. = Fall time.

D = Xo*t. = 193m/s * 20.3s = 3909 m.

2. V=sqrt(Xo^2+Yo^2)=sqrt(193^2+58^2) = 202 m/s.

3. Vo*t + 0.5g*t^2 = 2010 m.
58*t + 4.9*t^2 = 2010.
4.9t^2 + 58t - 2010 = 0.
Use Quadratic Formula.
t = 15.2 s. = Fall time.

D = 193m/s * 15.2s = 2934 m.

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Henry, I love you!!!! thank you so so much

1. Why did the package go to a party? Because it wanted to have a blast! The horizontal distance from the release point to the impact point can be found using the equation d = v*t, where d is the distance, v is the horizontal velocity of the plane (193 m/s), and t is the time it takes for the package to reach the ground. To find the time, we can use the equation h = (1/2)*g*t^2, where h is the initial height (2010 m) and g is the acceleration due to gravity (9.8 m/s^2). Rearranging this equation, we get t = sqrt(2h/g). Plugging in the values, we have t = sqrt(2*2010/9.8) = 63.35 seconds. Now we can calculate the distance: d = 193 m/s * 63.35 s = 12245.55 meters. So the horizontal distance from the release point to the impact point is approximately 12245.55 meters.

2. Why did the package need a passport? Because it wanted to travel with a velocity v1 = 58 m/s! The magnitude of the total velocity of the package at the moment it is thrown can be found using the Pythagorean theorem. We have a horizontal velocity of 193 m/s and a downward velocity of 58 m/s. So the total velocity can be calculated as sqrt((193 m/s)^2 + (58 m/s)^2) = sqrt(37369 + 3364) = sqrt(40733) ≈ 201.81 m/s. Therefore, the magnitude of the total velocity of the package at the moment it is thrown as seen by an observer on the ground is approximately 201.81 m/s.

3. Why did the package join a cross-country race? Because it wanted to cover some horizontal distance! The horizontal distance traveled by this package can be found by multiplying the horizontal velocity of the plane (193 m/s) by the time it takes for the package to reach the ground. We've already calculated the time to be 63.35 seconds in the previous question. So the horizontal distance is d = 193 m/s * 63.35 s = 12245.55 meters. Therefore, the horizontal distance traveled by this package is approximately 12245.55 meters. Keep those packages running!

1. To find the horizontal distance from the release point to the impact point, we need to consider the time it takes for the package to reach the ground.

The plane is flying horizontally at a speed of 193 m/s, so the horizontal velocity of the package is also 193 m/s. Since there is no horizontal acceleration, the horizontal velocity remains constant throughout.

Next, we need to find the time it takes for the package to hit the ground. We can use the formula for the vertical motion of an object in free fall:

h = (1/2)gt^2

where h is the initial height (2010 m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time.

Solving for t, we have:

t^2 = (2h)/g
t^2 = (2 * 2010) / 9.8
t^2 = 410.2041
t ≈ 20.2539 s

Now, we can find the horizontal distance using the formula:

d = v * t

where d is the horizontal distance, v is the horizontal velocity (193 m/s), and t is the time (20.2539 s):

d = 193 m/s * 20.2539 s ≈ 3919.28 m

Therefore, the horizontal distance from the release point to the impact point is approximately 3919.28 m.

2. To find the magnitude of the total velocity of the package at the moment it is thrown as seen by an observer on the ground, we need to consider the vertical and horizontal velocities separately.

The vertical velocity is given as v1 = 58 m/s. Since the package is thrown downward, we consider it negative in this case.

The horizontal velocity remains the same as before, v = 193 m/s.

To find the total velocity, we can use the Pythagorean theorem:

Total velocity^2 = (horizontal velocity)^2 + (vertical velocity)^2

Total velocity = √((horizontal velocity)^2 + (vertical velocity)^2)

Total velocity = √(193^2 + (-58)^2)
Total velocity ≈ √(37249 + 3364)
Total velocity ≈ √40613
Total velocity ≈ 201.52 m/s

Therefore, the magnitude of the total velocity of the package at the moment it is thrown, as seen by an observer on the ground, is approximately 201.52 m/s.

3. To find the horizontal distance traveled by the second package, we need to consider the time it takes for the package to hit the ground.

The vertical velocity of the package is given as v1 = 58 m/s. Since the package is thrown downward, we consider it negative in this case.

Using the same formula as before, h = (1/2)gt^2, we can solve for t:

t^2 = (2h)/g
t^2 = (2 * 2010) / 9.8
t^2 = 410.2041
t ≈ 20.2539 s

Now, we can find the horizontal distance using the formula:

d = v * t

where d is the horizontal distance, v is the horizontal velocity (193 m/s), and t is the time (20.2539 s):

d = 193 m/s * 20.2539 s ≈ 3919.28 m

Therefore, the horizontal distance traveled by the second package is approximately 3919.28 m.

thank you poopoo man

Glad I could help!