Answer:
a) t=3.6 sec
b) [tex]\mid v \mid =38\ m/s[/tex]
Explanation:
Horizontal Motion
When an object is thrown horizontally with a speed v from a height h, it describes a curved trajectory ruled by a constant speed in the horizontal direction and a variable speed in the vertical direction, where the acceleration of gravity makes the object fall to the ground.
If we know the height h from which the object was launched, the time it takes to hit the ground is:
[tex]\displaystyle t=\sqrt{\frac {2h}{g}}[/tex]
The horizontal speed is always constant:
vx=vo
But the vertical speed depends on the time and acceleration of gravity:
[tex]vy=g\cdot t[/tex]
The magnitude of the velocity or final speed of the object is given by:
[tex]\mid v \mid =\sqrt{vx^2+vy^2}[/tex]
a.
The stone was kicked over the cliff with a speed of v0=14 m/s, and the height it was thrown from is h=63.5 m, thus:
[tex]\displaystyle t=\sqrt{\frac {2\cdot 63.5}{9.8}}[/tex]
t=3.6 sec
b.
The vertical speed is:
[tex]vy=9.8\cdot 3.6[/tex]
vy=35.3 m/s
The final speed is calculated below:
[tex]\mid v \mid =\sqrt{14^2+35.3^2}[/tex]
[tex]\mid v \mid =\sqrt{1,442.09}[/tex]
[tex]\boxed{\mathbf{\mid v \mid =38\ m/s}}[/tex]
