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Chapter 3: Problem 88

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

Expert verified
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

01

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)
02

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: \(t \approx 5.98 \mathrm{~s}\)
03

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, \(v_{fy} = 7.50 - (9.81)(5.98) \approx -51.10 \mathrm{~m} / \mathrm{s}\)
04

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.
05

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.

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Most popular questions from this chapter

Some rental cars have a GPS unit installed, which allows the rental car company to check where you are at all times and thus also know your speed at any time. One of these rental cars is driven by an employee in the company's lot and, during the time interval from 0 to \(10 \mathrm{~s}\), is found to have a position vector as a function of time of $$ \begin{aligned} \vec{r}(t)=&\left((24.4 \mathrm{~m})-t(12.3 \mathrm{~m} / \mathrm{s})+t^{2}\left(2.43 \mathrm{~m} / \mathrm{s}^{2}\right)\right.\\\ &\left.(74.4 \mathrm{~m})+t^{2}\left(1.80 \mathrm{~m} / \mathrm{s}^{2}\right)-t^{3}\left(0.130 \mathrm{~m} / \mathrm{s}^{3}\right)\right) \end{aligned} $$ a) What is the distance of this car from the origin of the coordinate system at \(t=5.00 \mathrm{~s} ?\) b) What is the velocity vector as a function of time? c) What is the speed at \(t=5.00 \mathrm{~s} ?\) Extra credit: Can you produce a plot of the trajectory of the car in the \(x y\) -plane?

An air-hockey puck has a model rocket rigidly attached to it. The puck is pushed from one corner along the long side of the \(2.00-\mathrm{m}\) long air- hockey table, with the rocket pointing along the short side of the table, and at the same time the rocket is fired. If the rocket thrust imparts an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) to the puck, and the table is \(1.00 \mathrm{~m}\) wide, with what minimum initial velocity should the puck be pushed to make it to the opposite short side of the table without bouncing off either long side of the table? Draw the trajectory of the puck for three initial velocities: \(vv_{\min } .\) Neglect friction and air resistance.

In ideal projectile motion, when the positive \(y\) -axis is chosen to be vertically upward, the \(y\) -component of the velocity of the object during the ascending part of the motion and the \(y\) -component of the velocity during the descending part of the motion are, respectively, a) positive, negative. c) positive, positive. b) negative, positive. d) negative, negative.

The acceleration due to gravity on the Moon is \(1.62 \mathrm{~m} / \mathrm{s}^{2},\) approximately a sixth of the value on Earth. For a given initial velocity \(v_{0}\) and a given launch angle \(\theta_{0}\) the ratio of the range of an ideal projectile on the Moon to the range of the same projectile on Earth, \(R_{\text {Moon }} / R_{\text {Farth }}\) will be a) \(6 \mathrm{~m}\) d) \(5 \mathrm{~m}\) b) \(3 \mathrm{~m}\) e) \(1 \mathrm{~m}\) c) \(12 \mathrm{~m}\)

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