Problem:
(a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s.
(b) How long would it take to reach the ground if it is thrown straight down with the same speed?
Solution:
Part A
Refer to the figure below.
The known values are: t=2.35\:\text{s}; y=0\:\text{m}; v_0=+8.00\:\text{m/s}; and a=-9.8\:\text{m/s}^2
Based on the given values, the formula that we shall use is
y=y_0+v_0t+\frac{1}{2}at^2Substituting the values, we have
\begin{align*}
y & =y_0+v_0t+\frac{1}{2}at^2 \\
0\: & =y_0+\left(8.00\:\text{m/s}\right)\left(2.35\:\text{s}\right)+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(2.35\:\text{s}\right)^2 \\
y_0 & =8.26\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}Therefore, the cliff is 8.26 meters high.
Part B
Refer to the figure below
The knowns now are: y=0\:\text{m}; y_0=8.26\:\text{m}; v_0=-8.00\:\text{m/s}; and a=-9.80\:\text{m/s}^2
Based on the given values, we can use the formula
y=y_0+v_0t+\frac{1}{2}at^2Substituting the values, we have
\begin{align*}
y & =y_0+v_0t+\frac{1}{2}at^2 \\
0\:\text{m} & =8.26\:\text{m}+\left(-8.00\:\text{m/s}\right)t+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)t^2 \\
4.9 t^2+8t-8.26 & =0 \\
\end{align*}
Using the quadratic formula to solve for the value of t, we have
\begin{align*}
t &=\frac{-8\pm \sqrt{\left(8\right)^2-4\left(4.9\right)\left(-8.26\right)}}{2\left(4.9\right)} \\
t &=0.717\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}College Physics Chapter 2 Problems
College Physics 2nd Edition Solutions Table of Contents
Chapter 1: Introduction: The Nature of Science and Physics
Chapter 3: Two-Dimensional Kinematics
Chapter 4: Dynamics: Force and Newton’s Law of Motion
Chapter 5: Further Applications of Newton’s Laws: Friction, Drag, and Elasticity
Chapter 6: Uniform Circular Motion and Gravitation
Chapter 7: Work, Energy, and Energy Resources
Chapter 8: Linear Momentum and Collisions
Chapter 10: Rotational Motion and Angular Momentum
Chapter 12: Fluid Dynamics and Its Biological and Medical Applications
Chapter 13: Temperature, Kinetic Theory, and the Gas Laws
Chapter 14: Heat and Heat Transfer Methods
Chapter 15: Thermodynamics
Chapter 16: Oscillatory Motion and Waves
Chapter 17: Physics of Hearing
Chapter 18: Electric Charge and Electric Field
Chapter 19: Electric Potential and Electric Field
Chapter 20:
Electric Current, Resistance, and Ohm’s Law
Chapter 21: Circuits and DC Instruments
Chapter 22: Magnetism
Chapter 23: Electromagnetic Induction, AC Circuits, and Electrical Technologies
Chapter 24: Electromagnetic Waves
Chapter 25: Geometric Optics
Chapter 26: Vision and Optical Instrument
Chapter 27: Wave Optics
Chapter 28: Special Relativity
Chapter 29: Introduction to Quantum Physics
Chapter 30: Atomic Physics
Chapter 31: Radioactivity and Nuclear Physics
Chapter 32: Medical Applications of Nuclear Physics
Chapter 33: Particle Physics
Chapter 34: Frontiers of Physics
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