Question

A blimp is ascending at the rate of \(7.50 \mathrm{~m} / \mathrm{s}\) at a height of \(80.0 \mathrm{~m}\) above the ground when a package is thrown from its cockpit horizontally with a speed of \(4.70 \mathrm{~m} / \mathrm{s}\). a) How long does it take for the package to reach the ground? b) With what velocity (magnitude and direction) does it hit the ground?

Short answer

2. Which formula was used to find the time it takes for the package to reach the ground? 3. What was the horizontal component of the final velocity? 4. What is the significance of the negative sign in the angle of the final velocity? 5. What is the final velocity and angle with which the package hits the ground?

Step-by-step solution

Identify the known variables

The variables that we know are: - Initial vertical velocity, \(v_y = 7.50 \mathrm{~m} / \mathrm{s}\) (the blimp is ascending at this rate) - Initial horizontal velocity, \(v_x = 4.70 \mathrm{~m} / \mathrm{s}\) (package is thrown horizontally) - Initial height, \(h = 80.0 \mathrm{~m}\) (above the ground) - Acceleration due to gravity, \(g = 9.81 \mathrm{~m} / \mathrm{s^2}\) (downward direction)

Find the time it takes for a package to reach the ground

To find the time it takes for the package to reach the ground, we need to consider its vertical motion. We can use the following kinematic equation: \(h = v_y \times t - \frac{1}{2}gt^2\) Since the package starts at an initial height, we can rearrange the equation to solve for time \(t\): \(t = \frac{(-v_y) \pm \sqrt{v_y^2 + 2gh}}{g}\) Plugging in the known values, we get: \(t = \frac{(-7.50) \pm \sqrt{(7.50)^2 + 2(9.81)(80.0)}}{9.81}\) Use only the positive value of time, because the negative value doesn't make physical sense in this context: \(t \approx 5.98 \mathrm{~s}\)

Find the horizontal and vertical components of the final velocity

Now, we will find the horizontal and vertical components of the final velocity using the time found in the previous step. The horizontal component of the velocity remains constant because there is no horizontal acceleration: \(v_{fx} = v_x = 4.70 \mathrm{~m} / \mathrm{s}\) For the vertical component of the final velocity, we use the following equation: \(v_{fy} = v_y - gt\) Plugging in the known values, \(v_{fy} = 7.50 - (9.81)(5.98) \approx -51.10 \mathrm{~m} / \mathrm{s}\)

Find the magnitude and direction of the final velocity

To find the magnitude of the final velocity, we can use the Pythagorean theorem: \(v_f = \sqrt{v_{fx}^2 + v_{fy}^2}\) \(v_f = \sqrt{(4.70)^2 + (-51.10)^2} \approx 51.27 \mathrm{~m} / \mathrm{s}\) To find the direction of the final velocity, we can use the following formula: \(\theta = \arctan{(\frac{v_{fy}}{v_{fx}})}\) \(\theta = \arctan{(\frac{-51.10}{4.70})} \approx -84.92^\circ\) The negative sign in the angle means that the package is heading downward.

Final answers

a) The time it takes for the package to reach the ground is approximately \(5.98 \mathrm{~s}\). b) The package hits the ground with a velocity of approximately \(51.27 \mathrm{~m} / \mathrm{s}\) at an angle of \(-84.92^\circ\) below the horizontal.