**Solution** **Set Up:** We use our idealized model of projectile motion, in which we assume level ground, ignore the effects of air resistance, and treat the football as a point particle. We place the origin of coordinates at the point where the ball is kicked. Then \( z_0 = y_0 = 0 \). \( v_{0x} = v_0 \cos 30.0^\circ = (20.0\, \text{m/s})(0.866) = 17.3\, \text{m/s} \), \( v_{0y} = v_0 \sin 30.0^\circ = (20.0\, \text{m/s})(0.500) = 10.0\, \text{m/s} \). **Solve:** We first ask when (i.e., at what value of \( t \)) the ball is at a height of 3.05 m above the ground; we then find the value of \( z \) at that time. When that value of \( z \) is equal to the distance \( d \), the ball is just barely passing over the crossbar. To find the time \( t \) when \( y = 3.05\, \text{m} \), we use the equation \( y = y_0 + v_{0y} t - \frac{1}{2} g t^2 \). Substituting numerical values, we obtain \[ 3.05\, \text{m} = (10.0\, \text{m/s}) t - (4.90\, \text{m/s}^2) t^2 \] This is a quadratic equation; to solve it, we first write it in standard form: \[ (4.90\, \text{m/s}^2) t^2 - (10.0\, \text{m/s}) t + 3.05\, \text{m} = 0 \] Then we use the quadratic formula. We get \[ t = \frac{24 \pm 19}{9}\, \text{s} = \frac{(10.0\, \text{m/s}) \pm \sqrt{(10.0\, \text{m/s})^2 - 4(4.90\, \text{m/s}^2)(3.05
**Solution** **Set Up:** We use our idealized model of projectile motion, in which we assume level ground, ignore the effects of air resistance, and treat the football as a point particle. We place the origin of coordinates at the point where the ball is kicked. Then \( z_0 = y_0 = 0 \). \( v_{0x} = v_0 \cos 30.0^\circ = (20.0\, \text{m/s})(0.866) = 17.3\, \text{m/s} \), \( v_{0y} = v_0 \sin 30.0^\circ = (20.0\, \text{m/s})(0.500) = 10.0\, \text{m/s} \). **Solve:** We first ask when (i.e., at what value of \( t \)) the ball is at a height of 3.05 m above the ground; we then find the value of \( z \) at that time. When that value of \( z \) is equal to the distance \( d \), the ball is just barely passing over the crossbar. To find the time \( t \) when \( y = 3.05\, \text{m} \), we use the equation \( y = y_0 + v_{0y} t - \frac{1}{2} g t^2 \). Substituting numerical values, we obtain \[ 3.05\, \text{m} = (10.0\, \text{m/s}) t - (4.90\, \text{m/s}^2) t^2 \] This is a quadratic equation; to solve it, we first write it in standard form: \[ (4.90\, \text{m/s}^2) t^2 - (10.0\, \text{m/s}) t + 3.05\, \text{m} = 0 \] Then we use the quadratic formula. We get \[ t = \frac{24 \pm 19}{9}\, \text{s} = \frac{(10.0\, \text{m/s}) \pm \sqrt{(10.0\, \text{m/s})^2 - 4(4.90\, \text{m/s}^2)(3.05
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Solution**
**Set Up:** We use our idealized model of projectile motion, in which we assume level ground, ignore the effects of air resistance, and treat the football as a point particle. We place the origin of coordinates at the point where the ball is kicked. Then \( z_0 = y_0 = 0 \).
\( v_{0x} = v_0 \cos 30.0^\circ = (20.0\, \text{m/s})(0.866) = 17.3\, \text{m/s} \),
\( v_{0y} = v_0 \sin 30.0^\circ = (20.0\, \text{m/s})(0.500) = 10.0\, \text{m/s} \).
**Solve:** We first ask when (i.e., at what value of \( t \)) the ball is at a height of 3.05 m above the ground; we then find the value of \( z \) at that time. When that value of \( z \) is equal to the distance \( d \), the ball is just barely passing over the crossbar.
To find the time \( t \) when \( y = 3.05\, \text{m} \), we use the equation \( y = y_0 + v_{0y} t - \frac{1}{2} g t^2 \). Substituting numerical values, we obtain
\[ 3.05\, \text{m} = (10.0\, \text{m/s}) t - (4.90\, \text{m/s}^2) t^2 \]
This is a quadratic equation; to solve it, we first write it in standard form:
\[ (4.90\, \text{m/s}^2) t^2 - (10.0\, \text{m/s}) t + 3.05\, \text{m} = 0 \]
Then we use the quadratic formula. We get
\[ t = \frac{24 \pm 19}{9}\, \text{s} = \frac{(10.0\, \text{m/s}) \pm \sqrt{(10.0\, \text{m/s})^2 - 4(4.90\, \text{m/s}^2)(3.05](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F32f016ed-6c2c-4620-8b90-e10fe36181cc%2F57094b31-9d44-45e3-aa87-d19cd961566f%2Fj3zlgb5.jpeg&w=3840&q=75)
Transcribed Image Text:**Solution**
**Set Up:** We use our idealized model of projectile motion, in which we assume level ground, ignore the effects of air resistance, and treat the football as a point particle. We place the origin of coordinates at the point where the ball is kicked. Then \( z_0 = y_0 = 0 \).
\( v_{0x} = v_0 \cos 30.0^\circ = (20.0\, \text{m/s})(0.866) = 17.3\, \text{m/s} \),
\( v_{0y} = v_0 \sin 30.0^\circ = (20.0\, \text{m/s})(0.500) = 10.0\, \text{m/s} \).
**Solve:** We first ask when (i.e., at what value of \( t \)) the ball is at a height of 3.05 m above the ground; we then find the value of \( z \) at that time. When that value of \( z \) is equal to the distance \( d \), the ball is just barely passing over the crossbar.
To find the time \( t \) when \( y = 3.05\, \text{m} \), we use the equation \( y = y_0 + v_{0y} t - \frac{1}{2} g t^2 \). Substituting numerical values, we obtain
\[ 3.05\, \text{m} = (10.0\, \text{m/s}) t - (4.90\, \text{m/s}^2) t^2 \]
This is a quadratic equation; to solve it, we first write it in standard form:
\[ (4.90\, \text{m/s}^2) t^2 - (10.0\, \text{m/s}) t + 3.05\, \text{m} = 0 \]
Then we use the quadratic formula. We get
\[ t = \frac{24 \pm 19}{9}\, \text{s} = \frac{(10.0\, \text{m/s}) \pm \sqrt{(10.0\, \text{m/s})^2 - 4(4.90\, \text{m/s}^2)(3.05
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