an object is released off a 50m cliff. how long does it take to hit the ground and what is the final velocity of the object
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Answers
Explanation:
Free-Fall Motion:
Objects (or system of objects) that are either thrown vertically or fall downward exhibit motion that is affected by gravity. This motion, commonly referred as a free fall motion, is characterized as a rectilinear motion under the constant acceleration
g
=
−
9.8
m
/
s
2
that is always directed downward. Free-fall motions, just like horizontal rectilinear motions, are typically solved using appropriate kinematic formulas that relate the motion variables (displacement, time, velocity, acceleration) of an object.
Answer and Explanation:
The motion of the released object from the top of a cliff of height is that of a free-fall motion, which suggests that the time of drop can be determined from the kinematic formula:
y
=
v
0
y
t
+
1
2
g
t
2
,
wherein an expression for the time
t
given a zero initial velocity
v
0
y
=
0
m
/
s
is derived to be
t
=
√
2
y
g
Noting that the vertical displacement of the object's fall is
y
=
−
50
m
, then the time it takes for it to hit the ground is calculated to be
t
=
√
2
y
g
=
√
2
(
−
50
m
)
−
9.8
m
/
s
2
=
√
−
100
m
−
9.8
m
/
s
2
=
√
10.2
t
=
3.2
s
Hence, the released object will hit the ground at
t
=
3.2
s
Meanwhile, the object's final velocity (or the velocity as the object hits the ground) is calculated from the other kinematic equation:
v
y
=
v
0
+
g
t
2
=
g
t
2
Using the calculated value of
t
, then
v
y
=
g
t
=
(
−
9.8
m
/
s
2
)
(
3.2
s
)
v
y
=
−
31
m
/
s
A negative velocity signifies that the direction of its motion as it hits the ground is downward. Therefore, the final velocity of the object is
v
y
=
−
31
m
/
s
.
Answer:
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