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

A rock is thrown at an angle \(45^{\circ}\) below the horizontal from the top of a building. Immediately after release will its acceleration be greater than, equal to, or less than the acceleration due to gravity?

Short Answer

Expert verified
Answer: Immediately after release, the rock's acceleration will be equal to the acceleration due to gravity, which is approximately 9.81 m/sĀ².

Step by step solution

01

Identify the acceleration due to gravity

The acceleration due to gravity is a constant value and is always acting downwards. It's denoted as \(g\) and its value is approximately \(9.81 m/s^2\).
02

Calculate the initial horizontal and vertical velocities

The rock is thrown at an angle of \(45^{\circ}\) below the horizontal. We can split the initial velocity vector into its horizontal (\(v_x\)) and vertical (\(v_y\)) components. Assuming that \(V_0\) is the initial velocity of the rock, we have: \begin{align} v_x = V_0\cos(45^{\circ}) \\ v_y = -V_0\sin(45^{\circ}) \end{align} Notice that the vertical component is negative, as the angle is below the horizontal.
03

Examine the acceleration components

Now let's examine the horizontal and vertical components of the acceleration. In the horizontal direction, there is no force acting on the rock (ignoring air resistance), meaning that there is no acceleration in the horizontal direction: \begin{align} a_x = 0 \end{align} In the vertical direction, the only force acting on the rock is gravity, which causes the acceleration to have the same magnitude as gravity but in the opposite direction: \begin{align} a_y = -g \end{align}
04

Determine the total acceleration of the rock

The total acceleration of the rock is a vector with both horizontal and vertical components. To find this vector, we combine the horizontal and vertical components: \begin{align} \bold{a} = \bold{a_x} + \bold{a_y} \end{align} Since \(a_x = 0\), the total acceleration is: \begin{align} \bold{a} = \bold{a_y} \end{align} which means that the total acceleration of the rock is equal to the acceleration due to gravity. So, immediately after release, the rock's acceleration will be equal to the acceleration due to gravity.

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!

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

For a science fair competition, a group of high school students build a kicker-machine that can launch a golf ball from the origin with a velocity of \(11.2 \mathrm{~m} / \mathrm{s}\) and initial angle of \(31.5^{\circ}\) with respect to the horizontal. a) Where will the golf ball fall back to the ground? b) How high will it be at the highest point of its trajectory? c) What is the ball's velocity vector (in Cartesian components) at the highest point of its trajectory? d) What is the ball's acceleration vector (in Cartesian components) at the highest point of its trajectory?

Two cannonballs are shot in sequence from a cannon, into the air, with the same muzzle velocity, at the same launch angle. Based on their trajectory and range, how can you tell which one is made of lead and which one is made of wood. If the same cannonballs where launched in vacuum, what would the answer be?

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?

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 ).

A football is punted with an initial velocity of \(27.5 \mathrm{~m} / \mathrm{s}\) and an initial angle of \(56.7^{\circ} .\) What is its hang time (the time until it hits the ground again)?

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