A rubber ball is shot straight up from the ground with speed v₀. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?

Verified step by step guidance

1. Define the motion equations for both balls. For the first ball (shot upwards), its position as a function of time is given by: $x\_1(t)\; =\; v\_0\; t\; -\; \backslash (\backslash frac\{1\}\{2\}\backslash )\; g\; t^2$, where $v\_0$ is the initial velocity and $g$ is the acceleration due to gravity. For the second ball (dropped from height $h$), its position as a function of time is: $x\_2(t)\; =\; h\; -\; \backslash (\backslash frac\{1\}\{2\}\backslash )\; g\; t^2$.
2. Set the condition for collision. A collision occurs when the two balls are at the same position at the same time. This means $x\_1(t)\; =\; x\_2(t)$. Substitute the expressions for $x\_1(t)$ and $x\_2(t)$ into this equation: $v\_0\; t\; -\; \backslash (\backslash frac\{1\}\{2\}\backslash )\; g\; t^2\; =\; h\; -\; \backslash (\backslash frac\{1\}\{2\}\backslash )\; g\; t^2$.
3. Simplify the equation. Cancel out the common term $-\backslash (\backslash frac\{1\}\{2\}\backslash )\; g\; t^2$ on both sides to get: $v\_0\; t\; =\; h$. Rearrange this equation to solve for $h$: $h\; =\; v\_0\; t$.
4. Determine the maximum time $t$ for which the first ball is still moving upwards. The first ball reaches its highest point when its velocity becomes zero. Use the kinematic equation for velocity: $v\; =\; v\_0\; -\; g\; t$. Set $v\; =\; 0$ to find the time to reach the highest point: $t\; =\; \backslash (\backslash frac\{v\_0\}\{g\}\backslash )$.
5. Substitute the maximum time $t\; =\; \backslash (\backslash frac\{v\_0\}\{g\}\backslash )$ into the equation $h\; =\; v\_0\; t$ to find the maximum height $h$ for which a collision occurs: $h\; =\; v\_0\; \backslash (\backslash cdot\backslash )\; \backslash (\backslash frac\{v\_0\}\{g\}\backslash )\; =\; \backslash (\backslash frac\{v\_0^2\}\{g\}\backslash )$. This is the maximum value of $h$ for which the two balls collide before the first ball falls back to the ground.

Key Concepts

Kinematics

Kinematics is the branch of mechanics that describes the motion of objects without considering the forces that cause the motion. It involves concepts such as displacement, velocity, and acceleration. In this scenario, kinematic equations can be used to determine the positions of both balls over time, which is essential for analyzing their potential collision.

Free Fall

Free fall refers to the motion of an object under the influence of gravity alone, with no other forces acting on it. For the dropped ball, it accelerates downward at a constant rate of approximately 9.81 m/s². Understanding free fall is crucial for calculating how long it takes for the second ball to reach the ground and how high it can be dropped from to ensure a collision with the first ball.

Conservation of Energy

The conservation of energy principle states that energy cannot be created or destroyed, only transformed from one form to another. In this context, the kinetic energy of the first ball when shot upwards will convert to potential energy at its peak height. This concept helps in determining the maximum height the first ball reaches, which is necessary to find the maximum height h for the second ball to ensure a collision.