Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Short Answer

Expert verified
Milada lands 2.565 meters from the base of the cliff.

Step by step solution

01

Recognize the problem type

The problem involves projectile motion for Milada, who jumps horizontally off a cliff. We need to calculate how far from the base of the cliff Milada lands.
02

Understand Milada's vertical motion

Both crickets fall for 2.70 seconds since they start from the same height. This time will be used to calculate the horizontal distance Milada travels.
03

Set up horizontal motion equation

For horizontal motion, use the formula:\[x = v_0 imes t\]where \(x\) is the horizontal distance, \(v_0\) is the initial horizontal speed (95.0 cm/s, convert to meters: 0.95 m/s), and \(t\) is the time (2.70 s).
04

Calculate horizontal distance

Substitute the values into the horizontal motion equation:\[x = 0.95 \, \text{m/s} \times 2.70 \, \text{s} = 2.565 \, \text{m}\]Thus, Milada lands 2.565 meters from the base of the cliff.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Horizontal Motion Equation
In projectile motion, horizontal motion is straightforward. The key is to remember that horizontal velocity remains constant if air resistance is ignored. This is applicable when an object is projected horizontally from a height, like Milada the cricket. To find how far Milada lands from the cliff, we utilize the horizontal motion equation:
  • \( x = v_0 \times t \)
Here, \( x \) is the horizontal distance traveled, \( v_0 \) is the initial horizontal speed, and \( t \) is the time of flight.
Since there are no horizontal forces acting on Milada, the velocity and hence the initial speed (\( v_0 \)) stays the same throughout the flight.
This equation is essential for analyzing horizontal motion in projectile problems and, when applied correctly, allows us to calculate how far an object will land from its starting point on a horizontal path.
Initial Speed Conversion
Before using the horizontal motion equation, it's important to ensure unit consistency.
Milada's initial speed is given as 95.0 cm/s, which needs to be converted to meters per second for appropriate unit alignment in the equation:
  • Conversion factor: 1 cm = 0.01 m
  • Therefore, 95.0 cm/s × 0.01 = 0.95 m/s
This conversion ensures that every parameter in the formula \( x = v_0 \times t \) uses the metric system, providing an accurate computation of the horizontal distance.
Time of Flight
Time of flight is crucial in determining how far a projectile travels horizontally.
In this problem, both Chirpy and Milada experience the same time of flight – 2.70 seconds – as they fall from the cliff. This is because they both start their motion from the same height and are subject to gravity.
Understanding the time of flight involves analyzing the vertical motion aspect of projectile motion, which is influenced only by gravitational acceleration when air resistance is negligible.
  • Gravity doesn't affect horizontal motion directly.
  • Thus, once the time of flight is known from vertical analysis, it can be applied in the horizontal motion equation:
  • Utilize \( x = v_0 \times t \) with \( t = 2.70 \) seconds
This knowledge lets us calculate that Milada impacts the ground 2.565 meters from the base of the cliff, primarily dictated by the time she spends in the air.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0\(^\circ\) above the horizontal. The firefighters want to direct the water at a blaze that is 10.0 m above ground level. How far from the building should they position their cannon? There are \(two\) possibilities; can you get them both? (\(Hint\): Start with a sketch showing the trajectory of the water.)

The froghopper, \(Philaenus\) \(spumarius\), holds the world record for insect jumps. When leaping at an angle of 58.0\(^\circ\) above the horizontal, some of the tiny critters have reached a maximum height of 58.7 cm above the level ground. (See \(Nature\), Vol. 424, July 31, 2003, p. 509.) (a) What was the takeoff speed for such a leap? (b) What horizontal distance did the froghopper cover for this world-record leap?

A 2.7-kg ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 45.0-m-high cliff. At the instant the ball is thrown, a woman starts running away from the base of the cliff with a constant speed of 6.00 m/s. The woman runs in a straight line on level ground. Ignore air resistance on the ball. (a) At what angle above the horizontal should the ball be thrown so that the runner will catch it just before it hits the ground, and how far does she run before she catches the ball? (b) Carefully sketch the ball's trajectory as viewed by (i) a person at rest on the ground and (ii) the runner.

An airplane is flying with a velocity of 90.0 m/s at an angle of 23.0\(^{\circ}\) above the horizontal. When the plane is 114 m directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment. How far from the dog will the suitcase land? Ignore air resistance.

A rocket is fired at an angle from the top of a tower of height \(h_0\) = 50.0 m. Because of the design of the engines, its position coordinates are of the form \(x(t) = A + Bt^2 \)and \(y(t) = C + Dt^3\), where \(A, B, C,\) and \(D\) are constants. The acceleration of the rocket 1.00 s after firing is \(\vec{a} = (4.00 \hat{i}+ 3.00\hat{j}) m/s^2\). Take the origin of coordinates to be at the base of the tower. (a) Find the constants \(A, B, C,\) and \(D\), including their SI units. (b) At the instant after the rocket is fired, what are its acceleration vector and its velocity? (c) What are the \(x\)- and \(y\)-components of the rocket's velocity 10.0 s after it is fired, and how fast is it moving? (d) What is the position vector of the rocket 10.0 s after it is fired?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free