Respuesta :
Answer:
a) Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds, b) The initial velocity of the second stone is -16.038 meters per second, c) The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.
Explanation:
a) The time after the release after the release of the first stone can be get from the following kinematic formula for the first rock:
[tex]y_{1} = y_{1,o} + v_{1,o} \cdot t +\frac{1}{2}\cdot g \cdot t^{2}[/tex]
Where:
[tex]y_{1}[/tex] - Final height of the first stone, measured in meters.
[tex]y_{1,o}[/tex] - Initial height of the first stone, measured in meters.
[tex]v_{1,o}[/tex] - Initial speed of the first stone, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]g[/tex] - Gravity constant, measured in meters per square second.
Given that [tex]y_{1,o} = 47\,m[/tex], [tex]y_{1} = 0\,m[/tex], [tex]v_{1,o} = -2.12\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the following second-order polynomial is built:
[tex]-4.984\cdot t^{2} - 2.12\cdot t + 47 = 0[/tex]
Roots of the polynomial are, respectively:
[tex]t_{1} \approx 2.866\,s[/tex] and [tex]t_{2}\approx -3.291\,s[/tex]
Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds.
b) As the second stone is thrown a second later than first one, its height is represented by the following kinematic expression:
[tex]y_{2} = y_{2,o} + v_{2,o}\cdot (t-t_{o}) + \frac{1}{2}\cdot g \cdot (t-t_{o})^{2}[/tex]
[tex]y_{2}[/tex] - Final height of the second stone, measured in meters.
[tex]y_{2,o}[/tex] - Initial height of the second stone, measured in meters.
[tex]v_{2,o}[/tex] - Initial speed of the second stone, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]t_{o}[/tex] - Initial absolute time, measured in seconds.
[tex]g[/tex] - Gravity constant, measured in meters per square second.
Given that [tex]y_{2,o} = 47\,m[/tex], [tex]y_{2} = 0\,m[/tex], [tex]t_{o} = 1\,s[/tex], [tex]t = 2.866\,s[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the following expression is constructed and the initial speed of the second stone is:
[tex]1.866\cdot v_{2,o}+29.926 = 0[/tex]
[tex]v_{2,o} = -16.038\,\frac{m}{s}[/tex]
The initial velocity of the second stone is -16.038 meters per second.
c) The final speed of each stone is determined by the following expressions:
First stone
[tex]v_{1} = v_{1,o} + g \cdot t[/tex]
Second stone
[tex]v_{2} = v_{2,o} + g\cdot (t-t_{o})[/tex]
Where:
[tex]v_{1,o}, v_{1}[/tex] - Initial and final velocities of the first stone, measured in meters per second.
[tex]v_{2,o}, v_{2}[/tex] - Initial and final velocities of the second stone, measured in meters per second.
If [tex]v_{1,o} = -2.12\,\frac{m}{s}[/tex] and [tex]v_{2,o} = -16.038\,\frac{m}{s}[/tex], the final speeds of both stones are:
First stone
[tex]v_{1} = -2.12\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (2.866\,s)[/tex]
[tex]v_{1} = -30.227\,\frac{m}{s}[/tex]
Second stone
[tex]v_{2} = -16.038\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (2.866\,s-1\,s)[/tex]
[tex]v_{2} = -34.338\,\frac{m}{s}[/tex]
The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.
