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Crickets Chirpy and Milada jump from the top of a vertical cliff. Chirpy drops downward and reaches the ground in \(2.70 \mathrm{~s},\) while Milada jumps horizontally with an initial speed of \(95.0 \mathrm{~cm} / \mathrm{s}\). How far from the base of the cliff will Milada hit the ground? Ignore air resistance.

Short Answer

Expert verified
Milada will hit the ground \(2.565 m\) away from the base of the cliff.

Step by step solution

01

Determine the time it takes for the crickets to fall

Regardless of horizontal motion, both crickets should hit the ground at the same time as they are both falling from the same height. The time in free fall can be defined by the equation \(t = \sqrt{(2h/g)}\), where \(h\) is the height and \(g\) is the acceleration due to gravity. We're already given that Chirpy takes \(2.70 s\) to hit the ground, so this is the time for the fall.
02

Convert Units

Convert Milada's initial speed from \(cm/s\) to \(m/s\) for consistency with standard units in physics. The conversion is \(1 m = 100 cm\), so \(95.0 cm/s = 0.95 m/s\).
03

Calculate the horizontal distance

Use the time of the fall to calculate the horizontal distance Milada travels. The equation for the horizontal distance is \(d = vt\), where \(v\) is the velocity (or speed in this case) and \(t\) is the time. Substituting the given values, \(d = 0.95 m/s * 2.70 s = 2.565 m\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Free Fall
Imagine dropping an object from a certain height without any initial velocity—this is known as free fall. In free fall motion, only the force of gravity acts on the object, pulling it toward the Earth's surface with a constant acceleration. The remarkable thing about free fall is that all objects, regardless of their mass, experience the same acceleration due to gravity, which on Earth is approximately \( 9.8 \text{ m/s}^2 \). This is why our cricket pals, Chirpy and Milada, would hit the ground at the same time if Milada didn't jump off the cliff but simply dropped down like Chirpy.

Understanding free fall is crucial in solving a range of physics problems: it forms the basis of projectile motion and influences how objects move under the sole influence of gravity. In our example with the crickets, knowing that Chirpy takes 2.70 seconds to reach the ground allows us to calculate the height from which both crickets started their journey—essential information when dealing with projectile motion.
Constant Acceleration
In physics, constant acceleration occurs when an object's velocity changes by the same amount each second. An object in free fall, like Chirpy, is in constant acceleration due to gravity's unchanging rate. The equations used to describe motion—known as the kinematic equations—often rely on the presence of constant acceleration to predict the position and velocity of objects over time.

When acceleration is constant, the motion of the object can be easily predicted, and calculations become much more straightforward. This concept is essential in our cricket example since it allows us to accurately describe Chirpy's vertical journey regardless of the horizontal motion imparted to Milada.
Horizontal Projectile
A horizontal projectile, like Milada, involves an object moving with an initial horizontal velocity and simultaneously falling under gravity. In contrast to Chirpy, Milada experiences two components of motion: a constant horizontal velocity and a vertically accelerating motion due to gravity. When examining horizontal projectiles, it is critical to remember that the horizontal and vertical motions are independent of each other.

Milada's horizontal distance traveled relies solely on her initial horizontal speed and the time spent in the air, both of which remain uninfluenced by gravity. Calculating how far Milada lands from the cliff involves understanding that the horizontal motion is constant (assuming no air resistance)—emphasizing the importance of compartmentalizing the different aspects of projectile motion.
Kinematic Equations
Kinematic equations allow us to describe the motion of objects when acceleration is constant, such as during free fall or when an object moves under the force of gravity alone. These equations reliably connect variables like initial velocity, final velocity, acceleration, time, and displacement.

One of the primary kinematic equations used is \( s = ut + \frac{1}{2}at^2 \), where \( s \) is the displacement, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time. Chirpy's descent can be described by this equation with \( u = 0 \), as there is no initial vertical velocity and \( a = 9.8 \text{ m/s}^2 \) for the acceleration due to gravity. Milada's horizontal motion, despite being a projectile, can also be summed up by a simplified kinematic equation, \( d = vt \), since her horizontal acceleration is zero—highlighting the versatility of these formulas in tackling various motion problems.

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Most popular questions from this chapter

A squirrel has \(x\) - and \(y\) -coordinates \((1.1 \mathrm{~m}, 3.4 \mathrm{~m})\) at time \(t_{1}=0\) and coordinates \((5.3 \mathrm{~m},-0.5 \mathrm{~m})\) at time \(t_{2}=3.0 \mathrm{~s}\). For this time interval, find (a) the components of the average velocity, and (b) the magnitude and direction of the average velocity.

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A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by \(\overrightarrow{\boldsymbol{v}}=\) \(\left[5.00 \mathrm{~m} / \mathrm{s}-\left(0.0180 \mathrm{~m} / \mathrm{s}^{3}\right) t^{2}\right] \hat{\imath}+\left[2.00 \mathrm{~m} / \mathrm{s}+\left(0.550 \mathrm{~m} / \mathrm{s}^{2}\right) t\right] \hat{\jmath}\) (a) What are \(a_{x}(t)\) and \(a_{y}(t),\) the \(x\) - and \(y\) -components of the car's velocity as functions of time? (b) What are the magnitude and direction of the car's velocity at \(t=8.00 \mathrm{~s} ?\) (b) What are the magnitude and direction of the car's acceleration at \(t=8.00 \mathrm{~s} ?\)

A baseball thrown at an angle of \(60.0^{\circ}\) above the horizontal strikes a building \(18.0 \mathrm{~m}\) away at a point \(8.00 \mathrm{~m}\) above the point from which it is thrown. Ignore air resistance. (a) Find the magnitude of the ball's initial velocity (the velocity with which the ball is thrown). (b) Find the magnitude and direction of the velocity of the ball just before it strikes the building.

A railroad flatcar is traveling to the right at a speed of \(13.0 \mathrm{~m} / \mathrm{s}\) relative to an observer standing on the ground. Someone is riding a motor scooter on the flatcar (Fig. E3.36). What is the velocity (magnitude and direction) of the scooter relative to the flatcar if the scooter's velocity relative to the observer on the ground is (a) \(18.0 \mathrm{~m} / \mathrm{s}\) to the right? (b) \(3.0 \mathrm{~m} / \mathrm{s}\) to the left? (c) zero?

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