Question

You throw a glob of putty straight up toward the ceiling, which is 3.60 m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.50 m/s. (a) What is the speed of the putty just before it strikes the ceiling? (b) How much time from when it leaves your hand does it take the putty to reach the ceiling?

Short answer

(a) 4.43 m/s; (b) 0.517 s.

Step-by-step solution

Determine Final Velocity

We can use the kinematic equation \( v^2 = u^2 + 2as \) to find the final velocity of the putty just before it strikes the ceiling. Here, \( v \) is the final velocity, \( u = 9.50 \, \text{m/s} \) is the initial velocity, \( a = -9.81 \, \text{m/s}^2 \) is the acceleration due to gravity (negative because it's acting downward), and \( s = 3.60 \, \text{m} \) is the displacement. Substitute the values into the equation:\[ v^2 = (9.50)^2 + 2(-9.81)(3.60) \]\[ v^2 = 90.25 - 70.632 \]\[ v^2 = 19.618 \]\[ v = \sqrt{19.618} = 4.43 \, \text{m/s} \]Hence, the speed of the putty just before it strikes the ceiling is 4.43 m/s.

Calculate Time to Reach Ceiling

To find the time taken for the putty to reach the ceiling, we use the kinematic equation \( v = u + at \), where \( v \) is the final velocity from the previous step, \( u = 9.50 \, \text{m/s} \), \( a = -9.81 \, \text{m/s}^2 \), and \( t \) is the time. Rearrange to solve for \( t \):\[ t = \frac{v - u}{a} \]Substitute the values:\[ t = \frac{4.43 - 9.50}{-9.81} \]\[ t = \frac{-5.07}{-9.81} \]\[ t = 0.517 \, \text{s} \]Therefore, the putty takes approximately 0.517 seconds to reach the ceiling.

Key concepts

Projectile Motion

Projectile motion is the motion experienced by an object thrown into the air, subject to only the acceleration due to gravity. When a projectile is launched, it follows a curved path under the influence of gravity alone. In our scenario, the glob of putty is thrown upwards, making its motion an example of one-dimensional projectile motion.

There are a few characteristics of projectile motion:

• The motion has both vertical and horizontal components, but in this example, we only consider the vertical component since the putty moves straight up.
• The vertical motion behaves uniformly due to gravity acting as a constant downward force.
• At the peak of its motion (just before hitting the ceiling), the vertical velocity momentarily becomes zero before gravity pulls the object down again.

Understanding these components helps in analyzing how the putty behaves from the moment it leaves your hand until it strikes the ceiling.

Acceleration Due to Gravity

The acceleration due to gravity is a key factor in any projectile motion scenario. It acts as a constant force pulling the object back towards the Earth. This acceleration is symbolized as "g," and its standard value at the Earth's surface is approximately \(-9.81 \, \text{m/s}^2\) (the negative sign indicates a downward force).

In the exercise at hand:

• "g" slows down the upward ascent of the putty until it halts its rise.
• Subsequently, it accelerates the putty towards the ceiling, contributing to its final velocity just before impact.
• The constant value of \(9.81 \, \text{m/s}^2\) allows us to use standard kinematic equations effectively in our calculations.

Proper comprehension of gravity's impact on a projectile is vital. It helps in predicting the behavior of any object in free fall or thrown vertically.

Kinematic Equations

Kinematic equations are essential tools for solving problems involving uniform acceleration vector motion, such as projectile motion. These equations connect different physical quantities like velocity, displacement, time, and acceleration.

Key kinematic equations include:

• \(v = u + at\)- Relates final velocity \(v\), initial velocity \(u\), acceleration \(a\), and time \(t\).
• \(v^2 = u^2 + 2as\)- Useful for finding the final velocity without knowing the time directly.
• \(s = ut + \frac{1}{2}at^2\)- Calculates displacement \(s\), which is integrated over time from initial conditions.

In the problem, the equation \(v^2 = u^2 + 2as\) helped determine the speed of the putty just before it hit the ceiling. The time was then calculated using \(v = u + at\). By mastering these equations, one can solve a broad range of kinematics-related problems effortlessly.