**Delivering a Package by Air**

Part A = 11.95 s

Part B = 1,389 m

Part C = 369 mph

**Solution Below:**

Part A

Express your answer numerically in seconds. Neglect air resistance.

All that matters here is the height from which the package is dropped. The formula is:

x = v_{i}t + 1/2at^{2}

700 = 0 * t + 1/2 * 9.8 * t^{2}

700 = 4.9 * t^{2}

t^{2} = 142.857

t = 11.95 s

11.95 s

Part B

Express the distance numerically in meters.

First of all, convert mph into meters per second. Mastering Physics will constantly try and catch you with trick questions like this. Be very careful of units. One mph = 0.44704 m/s, so 260 mph = 116.23 m/s.

Now, since the package will be falling for 11.95 seconds and we know the velocity is 260 m/s, just use the formula x = v_{i}t + 1/2at^{2}. There is no horizontal acceleration because we ignore air resistance:

x = v_{i}t + 1/2at^{2}

x = 116.23 * 11.95 + 1/2 * 0 * 11.95^2

x = 116.23 * 11.95

x = 1,398 m

1,389 m

Part C

_{f}of the package when it hits the ground?

Express your answer numerically in miles per hour.

We need both the x and y components of the velocity to solve this part. Since we’re ignoring air resistance, the x component will be 260 mph (116.23 m/s – see Part B). The y component can be found with the formula v = at. Since the time is 11.95 seconds and acceleration is 9.8 m/s^2, the y component of the velocity is 117.11 m/s. To find the velocity, just use the Pythagorean theorem:

v_{f} = sqrt(v_{x}^2 + v_{y}^2)

v_{f} = sqrt(116.23^2 + 117.11^2)

v_{f} = 165 m/s

Now convert to mph. One mph is 0.44704 m/s, so:

V_{f} = m/s / 0.44704

V_{f} = 165 / 0.44704

V_{f} = 369 mph

369 mph