A cargo plane is moving with a horizontal velocity of

vx = +225 m/s
at a height of
y = 920 m
above level ground as shown in the figure below when it releases a package. Ignoring air resistance, how much time will it take the package to reach the ground? (Express your answer to the nearest tenth of a second.)

For this question you're going to need a kinematic equation. Because you're solving for time, I would use d=Voy*t+0.5*a*t^2

The acceleration is 9.8 m/s^2 (making "down" positive) due to gravity.

There is no initial velocity (V) in the y direction.

You do not need to use the fact that Vox is 225 m/s.

So if d=Voy*t+0.5*a*t^2 first substitute the values that you know. You get 920=0.5*(9.8)*t^2

solve for t. You should get something like 13.7 seconds.

Hope that helps!

y=gt²/2

t=sqrt(2y/g) =sqrt(2•920/9.8)=13.7 s

Well, if I were to drop a package from that height, I'd be more concerned about the package hitting me on the head than how much time it takes to reach the ground!

But hey, let's do some math anyway. We know that the package is not being launched horizontally, so we can ignore the horizontal velocity. The only force acting on it is gravity, pulling it straight down.

Using the equation y = -0.5gt^2 + vy0t + y0, where y is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), t is the time, vy0 is the initial vertical velocity, and y0 is the initial height, we can solve for t.

Plugging in the values, we get:

920 = -0.5 * 9.8 * t^2 + 0 * t + 0

Simplifying, we have:

920 = -4.9t^2

Dividing both sides by -4.9, we get:

t^2 = -920 / -4.9

t^2 ≈ 187.755

Taking the square root of both sides, we find:

t ≈ √187.755

t ≈ 13.7 s

So, it will take approximately 13.7 seconds for the package to reach the ground. Just enough time for me to find a nice, cozy spot to watch it land!

To find the time it takes for the package to reach the ground, we can use the equation of motion for vertical motion:

y = y0 + v0y * t - (1/2) * g * t^2

where:
y = final height (0 m, since it reaches the ground)
y0 = initial height (920 m)
v0y = initial vertical velocity (0 m/s, since the package is released horizontally)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time

In this case, the initial height is given as 920 m and the vertical velocity is 0 m/s since the package is released horizontally. We can solve the equation for t:

0 = 920 + 0 * t - (1/2) * 9.8 * t^2
0 = 920 - 4.9 * t^2

Rearranging the equation:

4.9 * t^2 = 920
t^2 = 920 / 4.9
t^2 = 187.7551

Taking the square root of both sides:

t ≈ √187.7551
t ≈ 13.6869 s

Therefore, it will take approximately 13.7 seconds for the package to reach the ground.

To find the time it takes for the package to reach the ground, we can use the equations of motion for vertical motion. The horizontal velocity of the plane is not relevant in this context since we are only concerned with the vertical motion of the package.

Since the package is released from rest (initial vertical velocity, vy = 0 m/s), we can use the equation:

y = vy * t + (1/2) * a * t^2

Here, y represents the vertical displacement (920 m), vy is the vertical velocity, t is the time taken, and a is the acceleration due to gravity (-9.8 m/s^2).

Rearranging the equation, we get:

t^2 - (2 * vy / a) * t - (2 * y / a) = 0

Comparing this with the general form of a quadratic equation (at^2 + bt + c = 0), we can see that a = 1, b = -(2 * vy / a), and c = -(2 * y / a).

Since we know the acceleration due to gravity is approximately -9.8 m/s^2, we can substitute these values in and solve for t using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Substituting the values, we get:

t = (-(2 * vy / a) ± √((2 * vy / a)^2 - 4 * (2 * y / a))) / (2 * a)

Substituting the given values:

t = (-(2 * 0) ± √((2 * 0)^2 - 4 * (2 * 920) * (-9.8))) / (2 * (-9.8))

Simplifying further:

t = ± √((0)^2 - 4 * (2 * 920) * (-9.8)) / (-19.6)

Since time can't be negative, we can ignore the negative square root. Calculating the square root expression:

t = √(4 * 920 * 9.8) / 19.6

t = √(35984) / 19.6

t ≈ √35984 / 19.6

Using a calculator, we find:

t ≈ 11.99 s

Therefore, it will take approximately 12 seconds (to the nearest tenth of a second) for the package to reach the ground.