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Chapter 3: Problem 16

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/s².

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.

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Most popular questions from this chapter

The captain of a boat wants to travel directly across a river that flows due east with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\). He starts from the south bank of the river and heads toward the north bank. The boat has a speed of \(6.10 \mathrm{~m} / \mathrm{s}\) with respect to the water. What direction (in degrees) should the captain steer the boat? Note that \(90^{\circ}\) is east, \(180^{\circ}\) is south, \(270^{\circ}\) is west, and \(360^{\circ}\) is north.

An outfielder throws the baseball to first base, located \(80 \mathrm{~m}\) away from the fielder, with a velocity of \(45 \mathrm{~m} / \mathrm{s}\). At what launch angle above the horizontal should he throw the ball for the first baseman to catch the ball in \(2 \mathrm{~s}\) at the same height? a) \(50.74^{\circ}\) c) \(22.7^{\circ}\) e) \(12.6^{\circ}\) b) \(25.4^{\circ}\) d) \(18.5^{\circ}\)

In an arcade game, a ball is launched from the corner of a smooth inclined plane. The inclined plane makes a \(30.0^{\circ}\) angle with the horizontal and has a width of \(w=50.0 \mathrm{~cm}\) The spring-loaded launcher makes an angle of \(45.0^{\circ}\) with the lower edge of the inclined plane. The goal is to get the ball into a small hole at the opposite corner of the inclined plane. With what initial velocity should you launch the ball to achieve this goal?

A rabbit runs in a garden such that the \(x\) - and \(y\) components of its displacement as function of times are given by \(x(t)=-0.45 t^{2}-6.5 t+25\) and \(y(t)=0.35 t^{2}+8.3 t+34 .\) (Both \(x\) and \(y\) are in meters and \(t\) is in seconds.) a) Calculate the rabbit's position (magnitude and direction) at \(t=10 \mathrm{~s}\) b) Calculate the rabbit's velocity at \(t=10 \mathrm{~s}\). c) Determine the acceleration vector at \(t=10 \mathrm{~s}\).

An air-hockey puck has a model rocket rigidly attached to it. The puck is pushed from one corner along the long side of the \(2.00-\mathrm{m}\) long air- hockey table, with the rocket pointing along the short side of the table, and at the same time the rocket is fired. If the rocket thrust imparts an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) to the puck, and the table is \(1.00 \mathrm{~m}\) wide, with what minimum initial velocity should the puck be pushed to make it to the opposite short side of the table without bouncing off either long side of the table? Draw the trajectory of the puck for three initial velocities: \(vv_{\min } .\) Neglect friction and air resistance.
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