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 \(\mathrm{m} / \mathrm{s} ?\) (c) If \(v_{0}\) is greater than some value \(v_{\text { max }},\) 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 \(v_{\text { min. }}\) The value \(v_{\min }\) also has a simple physical interpretation. What is it?

Step by step solution

01

Define Variables and Equations

Let the initial velocity of the first ball be \( v_0 \) and let its initial height be \( h = 20 \) m. The second ball is dropped after 1 second from the same height. The acceleration due to gravity, \( g \), is \( 9.8 \) m/s². Use the equations of motion: \( h = v_0 t - 0.5gt^2 \) for the first ball and \( h = h_0 - 0.5g(t-1)^2 \) for the second ball, where \( h_0 \) is the initial height of the building.

02

Calculate Time of Flight for Second Ball

Determine the time \( t_2 \) it takes for the second ball to hit the ground. It starts from rest and falls from a height of 20 m. Using the equation \( 0 = 20 - 0.5g(t-1)^2 \), solve for \( t_2 \).

03

Calculate Initial Velocity for First Ball

Substitute \( t_2 \) into the equation for the first ball. Use \( 0 = v_0 t - 0.5gt^2 \) (with \( h = 0 \)) to solve for \( v_0 \).

04

Solve Part (a) for Building Height

To solve for the height \( h \) when \( v_0 \) is given, rearrange the equations to find the relation between \( v_0 \) and \( h \) and set both times equal.

05

Calculate and Interpret V_max

Determine \( v_{max} \) where no \( h \) exists for the balls to meet timing requirements. Here, \( v_{max} \) is such that increasing \( h \) doesn't change the order of descent.

06

Calculate and Interpret V_min

Determine \( v_{min} \) as the lowest speed where increasing \( h \) makes no timing match possible. \( v_{min} \) is the speed that when decreased, results in the first ball always hitting last.

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Key Concepts

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

Equations of Motion

Projectile motion problems often involve objects moving under the influence of gravity. To solve these problems, we use the equations of motion. These equations help us calculate various aspects of an object's trajectory. For this exercise, we need to determine the initial speed and timing for the balls hitting the ground from a height.

• The first equation we use is: \( h = v_0 t - 0.5gt^2 \). This describes the motion of the first ball, which is thrown up with initial speed \( v_0 \).
• The second equation: \( h = h_0 - 0.5g(t-1)^2 \) is for the second ball, which is dropped after 1 second.
• The key variables here are time \( t \), initial velocity \( v_0 \), height \( h \), and the gravitational constant \( g \), typically \( 9.8 \text{ m/s}^2 \).

By analyzing these equations, you can solve for unknowns like the initial velocity or the time it takes for the balls to hit the ground.

Initial Velocity

The initial velocity, \( v_0 \), is the speed at which the first ball is thrown upwards. It is a crucial factor in determining how high the ball will go and how long it will take to reach certain points in its trajectory.

• For the first ball to land at the same time as the second ball, we need to find the right \( v_0 \) such that the times to hit the ground are equal.
• You can calculate \( v_0 \) by setting the equations for the two balls equal and solving for the velocity.
• This affects other parameters like maximum height achieved, which can be derived using the equation \( v^2 = v_0^2 - 2gh \).

The initial velocity determines the entire motion trajectory for the upward-thrown ball.

Gravity Effect

Gravity plays a significant role in projectile motion. It constantly influences the motion, pulling objects downwards at a rate of \( 9.8 \text{ m/s}^2 \).

• In this exercise, gravity affects both balls—one thrown upwards and one dropped from rest.
• The gravitational pull will slow down the first ball as it travels upward and speed it up as it falls back down.
• For the second ball, gravity causes continuous acceleration, impacting the overall time it takes to reach the ground.

Understanding gravity's effect is essential to predicting and calculating the balls' motion accurately.

Motion Graphs

Graphing the motion of projectiles helps visualize and understand their trajectories over time. For our exercise, imagining the path of each ball can clarify their motion.

• A typical graph would feature time on the x-axis and height on the y-axis, plotting each ball's path.
• The first ball's curve starts at the roof level, rises, and falls back down due to gravity.
• The second ball's line begins after a delay, showing a straightforward descent due to being dropped.

Such graphs help identify the timings and interaction points where both balls hit the ground simultaneously.