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Chapter 3: Problem 80

Rain is falling vertically at a constant speed of \(7.00 \mathrm{~m} / \mathrm{s}\). At what angle from the vertical do the raindrops appear to be falling to the driver of a car traveling on a straight road with a speed of \(60.0 \mathrm{~km} / \mathrm{h} ?\)

Short Answer

Expert verified
Answer: The raindrops appear to be falling at an angle of 67.32° from the vertical to the driver of the car.

Step by step solution

01

Convert the car's speed to m/s

The speed of the car is given in km/h. We need to convert this to m/s in order to compare it directly with the raindrop's speed. To do this, we will use the conversion factor: 1 km = 1000 m and 1 h (hour) = 3600 s (seconds). So, the car's speed is: \(60.0 \mathrm{~km/h} * \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} * \frac{1 \mathrm{~h}}{3600 \mathrm{~s}} = 16.67 \mathrm{~m/s}\).
02

Calculate the relative speed of raindrops

Since the car is moving horizontally, and rain is falling vertically, the relative motion of raindrops will be the result of adding the horizontal and vertical velocity components: \(\text{Relative velocity (Vi)}\) of raindrops = \(\sqrt{(speed\, of\, car)^2 + (speed\, of\, raindrops)^2} = \sqrt{(16.67\, m/s)^2 + (7.00\, m/s)^2} = \sqrt{278.89\, m^2/s^2 + 49.00\, m^2/s^2} = \sqrt{327.89\, m^2/s^2} = 18.11\, m/s\)
03

Find the inclination angle

Now that we have the relative speed of the raindrops, let's find the angle at which they are falling relative to the car's motion. We can use the tangent function along with the horizontal and vertical speed components to solve for the angle (let's call it \(\theta\)): \(\tan(\theta) = \frac{speed\, of\, car}{speed\, of\, raindrops} = \frac{16.67\, m/s}{7.00\, m/s}\), Now, to find the angle \(\theta\), we take the inverse tangent (also called arctangent) of both sides: \(\theta = \arctan(\frac{16.67\, m/s}{7.00\, m/s}) = \arctan(2.38) \approx 67.32^{\circ}\). So, the raindrops appear to be falling at an angle of \(67.32^{\circ}\) from the vertical to the driver of the car.

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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?

A circus juggler performs an act with balls that he tosses with his right hand and catches with his left hand. Each ball is launched at an angle of \(75^{\circ}\) and reaches a maximum height of \(90 \mathrm{~cm}\) above the launching height. If it takes the juggler \(0.2 \mathrm{~s}\) to catch a ball with his left hand, pass it to his right hand and toss it back into the air, what is the maximum number of balls he can juggle?

On a battlefield, a cannon fires a cannonball up a slope, from ground level, with an initial velocity \(v_{0}\) at an angle \(\theta_{0}\) above the horizontal. The ground itself makes an angle \(\alpha\) above the horizontal \(\left(\alpha<\theta_{0}\right) .\) What is the range \(R\) of the cannonball, measured along the inclined ground? Compare your result with the equation for the range on horizontal ground (equation 3.25 ).

An outfielder throws a baseball with an initial speed of \(32 \mathrm{~m} / \mathrm{s}\) at an angle of \(23^{\circ}\) to the horizontal. The ball leaves his hand from a height of \(1.83 \mathrm{~m}\). How long is the ball in the air before it hits the ground?

For a given initial speed of an ideal projectile, there is (are) \(\quad\) launch angle(s) for which the range of the projectile is the same. a) only one b) two different c) more than two but a finite number of d) only one if the angle is \(45^{\circ}\) but otherwise two different e) an infinite number of
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