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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

A baseball is thrown with a velocity of \(31.1 \mathrm{~m} / \mathrm{s}\) at an angle of \(\theta=33.4^{\circ}\) above horizontal. What is the horizontal component of the ball's velocity at the highest point of the ball's trajectory?

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