91Ó°ÊÓ

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.00 s later. You may ignore air resistance. (a) If the height of the building is \(20.0 \mathrm{m},\) what must the initial speed of the first ball be if both are to hit the ground at the same time? On the same graph, sketch the position of each ball as a function of time, measured from when the first ball is thrown. Consider the same situation, but now let the initial speed \(v_{0}\) of the first ball be given and treat the height \(h\) of the building as an unknown. (b) What must the height of the building be for both balls to reach the ground at the same time (i) if \(v_{0}\) is 6.0 \(\mathrm{m} / \mathrm{s}\) and (ii) if \(v_{0}\) is 9.5 \(\mathrm{m} / \mathrm{s} ?(\mathrm{c})\) If \(v_{0}\) is greater than some value \(v_{\max }\) a value of \(h\) does not exist that allows both balls to hit the ground at the same time. Solve for \(v_{\text { max. The value }} v_{\text { max }}\) has a simple physical interpretation. What is it? (d) If \(v_{0}\) is less than some value \(v_{\text { min }}\) a value of \(h\) does not exist that allows both balls to hit the ground at the same time. Solve for \(v_{\text { main }}\) The value \(v_{\text { min }}\) also has a simple physical interpretation. What is it?

Step by step solution

01

Understanding the scenario

We have two balls: the first one is thrown upward and the second one is dropped after 1 second. We are to determine conditions such that both hit the ground simultaneously.

02

Equations of motion for ball 1 and ball 2

For ball 1: It is thrown upward with initial velocity \( v_0 \) from height \( h = 20 \text{ m} \). The equation of motion is given by \( y_1(t) = h + v_0 t - \frac{1}{2}gt^2 \). For ball 2: It is released after 1 second, so its motion starts from \( t = 1 \text{ s} \). The equation is \( y_2(t) = h - \frac{1}{2}g(t-1)^2 \).

03

Condition for simultaneous impact

Both balls hit the ground simultaneously, so \( y_1(t_1) = 0 \) and \( y_2(t_2) = 0 \) at the same time \( t_1 = t_2 \). Set the equations to zero to find \( t \) when they reach the ground: \[ 0 = 20 + v_0 t - \frac{1}{2}gt^2 \] \[ 0 = 20 - \frac{1}{2}g(t-1)^2 \]

04

Solve for initial velocity (Part a)

Given \( h = 20 \text{ m} \), solve the second equation: \[ 0 = 20 - \frac{1}{2}g(t_2-1)^2 \] This gives \( t_2 = \sqrt{\frac{40}{g}} + 1 \). Substitute \( t_2 \) into the first equation to solve for \( v_0 \).

05

Solve for building height (Part b)

To find height \( h \), rearrange the equations assuming \( v_0 = 6 \text{ m/s} \) or \( v_0 = 9.5 \text{ m/s} \), then solve:1. Replace \( t \) with \( t_2 \) from previous step.2. Calculate \( h \) using:\[ h + v_0 t - \frac{1}{2}gt^2 = 0 \] and \[ h - \frac{1}{2}g(t-1)^2 = 0 \].

06

Determine maximum initial velocity (Part c)

To find \( v_{\text{max}} \), observe the physical scenario: If the ball is thrown upward too fast, it won't descend to the level of the dropped ball in time. \( v_{\max} \) is when the first ball's time of flight equals the second ball's fall after the delay. Solve:1. Set \( t_2 - 1 = \frac{2v_{\text{max}}}{g} \).2. Solve for \( v_{\text{max}} = 10 \text{ m/s} \).

07

Determine minimum initial velocity (Part d)

If the first ball is thrown too slow, it won't reach the ground if both balls fall freely. For the threshold scenario where both hit simultaneously, solve:1. Set conditions such that both equations are solvable for positive \( t \).2. Calculate \( v_{\text{min}} = 0 \text{ m/s} \), indicating if the first ball is not thrown upward at all, it won't meet the condition.

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Key Concepts

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

Kinematics

In physics, kinematics focuses on the description of motion without considering the forces causing it. It's like describing how something moves based on its speed, direction, and position changes over time. When faced with a problem involving motion, kinematics helps us determine important details such as velocity and position during the motion.

In the exercise, the motion of two balls is described by their positions over time. Understanding kinematics allows us to predict where these balls are at any given moment and determine their paths.

• The first ball is thrown upwards, meaning it initially moves against gravity, slows down, stops momentarily, and then falls back down.
• The second ball is dropped, so it starts to accelerate downwards due to gravity immediately.

By understanding these principles, kinematics helps solve real-world problems like determining when and where objects land.

Equations of Motion

The equations of motion provide mathematical tools to describe an object's movement. For any body in uniform acceleration, such as falling objects under gravity, these equations help predict future behavior.

• The first equation of motion for the upward thrown ball is: \[ y_1(t) = h + v_0 t - \frac{1}{2}gt^2 \]
• For the second ball, which is released later, the equation modifies to account for its starting time: \[ y_2(t) = h - \frac{1}{2}g(t-1)^2 \]

In these equations,

• \(y(t)\) represents the position of the ball at time \(t\)
• \(v_0\) is the initial speed of the thrown ball,
• and \(g\) is the acceleration due to gravity – typically \(9.8 \, \text{m/s}^2 \).

By setting these equations equal to zero, we can solve for the times when each ball hits the ground. This approach helps us determine necessary conditions like initial speed and building height for simultaneous impact.

Initial Velocity

Initial velocity is the speed at which an object begins its motion. In projectile problems, initial velocity is critical to determining the motion's path and outcome.

• In the exercise, understanding the significance of initial velocity is key to ensuring both balls hit the ground simultaneously.
• For the first ball, it's thrown upwards with \(v_0\), which must have enough speed to allow it to reach the ground at the same time as the second ball that's simply dropped.

To find that initial speed, you solve the motion equations for a specific value of \(t\) that ensures both balls reach the ground at the same moment. Initial velocities of 6 m/s and 9.5 m/s are considered in the exercise, representing different problem conditions.

Free Fall

Free fall describes a condition where gravity is the only force acting on an object. Without air resistance, any dropped object will accelerate downwards at about \(9.8 \, \text{m/s}^2\). It's an idealized form of motion that simplifies problems.

• When the second ball starts its descent after the delay, it enters free fall, influenced solely by gravity.
• Similarly, once the first ball reaches its peak and starts descending, it also moves under free fall conditions.

In projectile motion exercises like this one, free fall is significant because it simplifies the calculations needed to determine when and where an object will hit the ground. By recognizing the free-fall phase, you can focus on equations of motion that rely solely on gravitational force, making predictions more straightforward.