Textbook Question
A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in s. You may ignore air resistance, so the brick is in free fall. How tall, in meters, is the building?
1
views
Ch 02: Motion Along a Straight Line
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch 01: Units, Physical Quantities & Vectors37
- Ch 02: Motion Along a Straight Line57
- Ch 03: Motion in Two or Three Dimensions45
- Ch 04: Newton's Laws of Motion23
- Ch 05: Applying Newton's Laws54
- Ch 06: Work & Kinetic Energy11
- Ch 07: Potential Energy & Conservation6
- Ch 08: Momentum, Impulse, and Collisions15
- Ch 09: Rotation of Rigid Bodies37
- Ch 10: Dynamics of Rotational Motion44
- Ch 11: Equilibrium & Elasticity27
- Ch 12: Fluid Mechanics27
- Ch 13: Gravitation34
- Ch 14: Periodic Motion26
- Ch 15: Mechanical Waves22
- Ch 16: Sound & Hearing28
- Ch 17: Temperature and Heat28
- Ch 18: Thermal Properties of Matter35
- Ch 19: The First Law of Thermodynamics29
- Ch 20: The Second Law of Thermodynamics18
- Ch 21: Electric Charge and Electric Field28
- Ch 22: Gauss' Law21
- Ch 23: Electric Potential31
- Ch 24: Capacitance and Dielectrics18
- Ch 25: Current, Resistance, and EMF24
- Ch 26: Direct-Current Circuits29
- Ch 27: Magnetic Field and Magnetic Forces16
- Ch 28: Sources of Magnetic Field10
- Ch 29: Electromagnetic Induction22
- Ch 30: Inductance14
- Ch 31: Alternating Current20
- Ch 33: The Nature and Propagation of Light17
- Ch 34: Geometric Optics29
- Ch 35: Interference13
- Ch 36: Diffraction19
- Ch 37: Special Relativity19
- Ch 38: Photons: Light Waves Behaving as Particles12
- Ch 39: Particles Behaving as Waves25
- Ch 40: Quantum Mechanics I: Wave Functions20
- Ch 41: Quantum Mechanics II: Atomic Structure21
- Ch 42: Molecules and Condensed Matter17
- Ch 43: Nuclear Physics13
- Ch 44: Particle Physics and Cosmology17
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 2, Problem 38b
A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in s. You may ignore air resistance, so the brick is in free fall. What is the magnitude of the brick's velocity just before it reaches the ground?
Verified step by step guidance1
Identify the known values: The initial velocity \( v_0 \) is 0 m/s, the time \( t \) is 1.90 s, and the acceleration due to gravity \( g \) is approximately 9.81 m/s².
Use the kinematic equation for velocity in free fall: \( v = v_0 + gt \). Since the initial velocity \( v_0 \) is zero, the equation simplifies to \( v = gt \).
Substitute the known values into the equation: \( v = 9.81 \text{ m/s}^2 \times 1.90 \text{ s} \).
Calculate the product of the acceleration due to gravity and the time to find the magnitude of the velocity just before the brick hits the ground.
The result from the calculation will give you the magnitude of the brick's velocity just before impact, which is the final answer.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
2mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Free Fall
Free fall refers to the motion of an object under the influence of gravitational force only, with no other forces acting on it, such as air resistance. In this scenario, the brick is in free fall, meaning it accelerates downward at a constant rate due to gravity, which is approximately 9.81 m/s² on Earth.
Recommended video:
Guided course
08:36
Vertical Motion & Free Fall
Acceleration Due to Gravity
Acceleration due to gravity is the rate at which an object speeds up as it falls freely towards the Earth. It is denoted by 'g' and has a standard value of 9.81 m/s². This constant acceleration is crucial for calculating the velocity and displacement of objects in free fall, such as the brick in this problem.
Recommended video:
Guided course
05:20Acceleration Due to Gravity
Kinematic Equations
Kinematic equations describe the motion of objects under constant acceleration. For this problem, the equation v = u + gt can be used, where v is the final velocity, u is the initial velocity (zero in this case), g is the acceleration due to gravity, and t is the time. This equation helps determine the brick's velocity just before it hits the ground.
Recommended video:
Guided course
08:25Kinematics Equations
Related Practice
Textbook Question
A tennis ball on Mars, where the acceleration due to gravity is and air resistance is negligible, is hit directly upward and returns to the same level s later. How high above its original point did the ball go?
2
views
Textbook Question
A -kg rocket blasts off vertically from the launch pad with a constant upward acceleration of m/s2 and feels no appreciable air resistance. When it has reached a height of m, its engines suddenly fail; the only force acting on it is now gravity. What is the maximum height this rocket will reach above the launch pad?
2
views
Textbook Question
A -kg rocket blasts off vertically from the launch pad with a constant upward acceleration of m/s2 and feels no appreciable air resistance. When it has reached a height of m, its engines suddenly fail; the only force acting on it is now gravity. How much time will elapse after engine failure before the rocket comes crashing down to the launch pad, and how fast will it be moving just before it crashes?
2
views
Textbook Question
A lunar lander is making its descent to Moon Base I (Fig. E). The lander descends slowly under the retro-thrust of its descent engine. The engine is cut off when the lander is m above the surface and has a downward speed of m/s. With the engine off, the lander is in free fall. What is the speed of the lander just before it touches the surface? The acceleration due to gravity on the moon is m/s.
3
views
Textbook Question
A tennis ball on Mars, where the acceleration due to gravity is and air resistance is negligible, is hit directly upward and returns to the same level s later. How fast was it moving just after it was hit?
2
views