Ch 02: Motion Along a Straight Line

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 02: Motion Along a Straight Line

Problem 34b

Chapter 2, Problem 34b

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. How much time from when it leaves your hand does it take the putty to reach the ceiling?

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Identify the known values: the initial velocity $ v_0 = 9.50 \text{ m/s} $, the displacement $ s = 3.60 \text{ m} $, and the acceleration due to gravity $ a = -9.81 \text{ m/s}^2 $ (negative because it acts downward).

Use the kinematic equation for displacement: $ s = v_0 t + \frac{1}{2} a t^2 $. Substitute the known values into this equation: $ 3.60 = 9.50 t + \frac{1}{2} (-9.81) t^2 $.

Rearrange the equation to form a quadratic equation: $ 0 = -4.905 t^2 + 9.50 t - 3.60 $.

Solve the quadratic equation using the quadratic formula: $ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a = -4.905 $, $ b = 9.50 $, and $ c = -3.60 $.

Calculate the discriminant $ b^2 - 4ac $ to ensure it is non-negative, then solve for $ t $ using the quadratic formula to find the time it takes for the putty to reach the ceiling.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Kinematics Equations

Kinematics equations describe the motion of objects without considering the forces that cause the motion. In this problem, the equation v = u + at can be used to find the time taken for the putty to reach the ceiling, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity, and t is the time.

Acceleration due to Gravity

Acceleration due to gravity is a constant force acting on objects near the Earth's surface, typically approximated as 9.81 m/s² downward. This concept is crucial for calculating the motion of the putty as it travels upward against gravity, slowing down until it reaches the ceiling.

Projectile Motion

Projectile motion involves the movement of an object thrown into the air, subject to gravitational acceleration. Understanding this concept helps in analyzing the vertical motion of the putty, as it is projected upwards and decelerates until it reaches its peak height at the ceiling.

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Related Practice

Textbook 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. What is the speed of the putty just before it strikes the ceiling?

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Textbook Question

A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in $1.90$ s. You may ignore air resistance, so the brick is in free fall. How tall, in meters, is the building?

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Textbook Question

A tennis ball on Mars, where the acceleration due to gravity is $0.379 g$ and air resistance is negligible, is hit directly upward and returns to the same level $8.5$ s later. How high above its original point did the ball go?

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Textbook Question

A small rock is thrown vertically upward with a speed of $22.0$ m/s from the edge of the roof of a $30.0$ -m-tall building. The rock doesn't hit the building on its way back down and lands on the street below. Ignore air resistance. How much time elapses from when the rock is thrown until it hits the street?

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Textbook Question

A juggler throws a bowling pin straight up with an initial speed of $8.20$ m/s. How much time elapses until the bowling pin returns to the juggler's hand?

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Textbook Question

A tennis ball on Mars, where the acceleration due to gravity is $0.379g$ and air resistance is negligible, is hit directly upward and returns to the same level $8.5$ s later. How fast was it moving just after it was hit?

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You throw a glob of putty straight up toward the ceiling, which is 3.603.60 m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.509.50 m/s. How much time from when it leaves your hand does it take the putty to reach the ceiling?

Verified video answer for a similar problem:

Key Concepts

Kinematics Equations

Acceleration due to Gravity

Projectile Motion

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. How much time from when it leaves your hand does it take the putty to reach the ceiling?