Answer:
a) The stone takes approximately 6.704 second to reach the ground.
b) The stone takes approximately 6.704 second to reach the ground.
c) The stone hits the ground at a velocity of -215.695 meters per second.
Step-by-step explanation:
a) The stone experiments a free fall, which is a uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation are negliglble. Then, the vertical distance of the stone above the ground at any time [tex]t[/tex] is represented by the following model:
[tex]y = y_{o}+v_{o}\cdot t +\frac{1}{2}\cdot g\cdot t^{2}[/tex] (1)
Where:
[tex]y_{o}[/tex] - Initial height of the stone, measured in meters.
[tex]v_{o}[/tex] - Initial velocity of the stone, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If we know that [tex]y_{o} = 723\,ft[/tex], [tex]v_{o} = 0\,\frac{ft}{s}[/tex] and [tex]g = -32.174\,\frac{ft}{s^{2}}[/tex], then the distance of the stone above the ground at any time [tex]t[/tex] is represented by the following model:
[tex]y = 723 -16.087\cdot t^{2}[/tex] (2)
b) If we know that [tex]y = 0[/tex], then the time taken by the stone to reach the ground is:
[tex]723-16.087\cdot t^{2} = 0[/tex]
[tex]t = \sqrt{\frac{723}{16.087} }\,s[/tex]
[tex]t \approx 6.704\,s[/tex]
The stone takes approximately 6.704 second to reach the ground.
c) The function velocity is found by differentiating (2) once:
[tex]v = -32.174\cdot t[/tex] (3)
If we know that [tex]t \approx 6.704\,s[/tex], then the velocity when it hits the ground is:
[tex]v = -32.174\cdot (6.704)[/tex]
[tex]v = -215.695\,\frac{m}{s}[/tex]
The stone hits the ground at a velocity of -215.695 meters per second.
