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Chapter 3: Problem 75

At the end of the spring term, a high school physics class celebrates by shooting a bundle of exam papers into the town landfill with a homemade catapult. They aim for a point that is \(30.0 \mathrm{~m}\) away and at the same height from which the catapult releases the bundle. The initial horizontal velocity component is \(3.90 \mathrm{~m} / \mathrm{s}\). What is the initial velocity component in the vertical direction? What is the launch angle?

Short Answer

Expert verified
Answer: The initial vertical velocity is 37.7 m/s, and the launch angle is approximately 84.2°.

Step by step solution

01

Calculate time of flight

Using the first equation: \(x = x_0 + v_{0x}t_{\text{flight}}\) Since the initial position \(x_0 = 0\), we can write: \(t_{\text{flight}} = \frac{x}{v_{0x}}\) Plugging in the given values, we get: \(t_{\text{flight}} = \frac{30.0 \mathrm{~m}}{3.90 \mathrm{~m} / \mathrm{s}} \approx 7.69 \mathrm{~s}\)
02

Calculate initial vertical velocity

Using the second equation: \(0 = v_{0y}t_{\text{flight}} - \frac{1}{2}gt^2_{\text{flight}}\) Solving for \(v_{0y}\), we get: \(v_{0y} = \frac{1}{2}gt_{\text{flight}}\) Using \(g = 9.81 \mathrm{~m} / \mathrm{s}^2\) and \(t_{\text{flight}} \approx 7.69 \mathrm{~s}\), we calculate \(v_{0y}\): \(v_{0y} = \frac{1}{2}(9.81 \mathrm{~m} / \mathrm{s}^2)(7.69 \mathrm{~s}) \approx 37.7 \mathrm{~m} / \mathrm{s}\)
03

Calculate launch angle

Using the third equations, we get \(\tan{\theta} = \frac{v_{0y}}{v_{0x}}\) Plugging in the known values: \(\tan{\theta} = \frac{37.7 \mathrm{~m} / \mathrm{s}}{3.90 \mathrm{~m} / \mathrm{s}} \approx 9.67\) Now, compute the launch angle: \(\theta = \arctan{(9.67)} \approx 84.2^\circ\) So, the initial vertical velocity component is \(37.7\mathrm{~m} /\mathrm{s}\), and the launch angle is approximately \(84.2^\circ\).

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Most popular questions from this chapter

10.0 seconds after being fired, a cannonball strikes a point \(500 . \mathrm{m}\) horizontally from and \(100 . \mathrm{m}\) vertically above the point of launch. a) With what initial velocity was the cannonball launched? b) What maximum height was attained by the ball? c) What is the magnitude and direction of the ball's velocity just before it strikes the given point?

In a projectile motion, the horizontal range and the maximum height attained by the projectile are equal. a) What is the launch angle? b) If everything else stays the same, how should the launch angle, \(\theta_{0},\) of a projectile be changed for the range of the projectile to be halved?

A projectile is launched at an angle of \(45^{\circ}\) above the horizontal. What is the ratio of its horizontal range to its maximum height? How does the answer change if the initial speed of the projectile is doubled?

In a proof-of-concept experiment for an antiballistic missile defense system, a missile is fired from the ground of a shooting range toward a stationary target on the ground. The system detects the missile by radar, analyzes in real time its parabolic motion, and determines that it was fired from a distance \(x_{0}=5000 \mathrm{~m}\), with an initial speed of \(600 \mathrm{~m} / \mathrm{s}\) at a launch angle \(\theta_{0}=20^{\circ} .\) The defense system then calculates the required time delay measured from the launch of the missile and fires a small rocket situated at \(y_{0}=500 \mathrm{~m}\) with an initial velocity of \(v_{0} \mathrm{~m} / \mathrm{s}\) at a launch angle \(\alpha_{0}=60^{\circ}\) in the \(y z\) -plane, to intercept the missile. Determine the initial speed \(v_{0}\) of the intercept rocket and the required time delay.

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?
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