Answer:
The time to take the marble to reach the floor is 0.553 s and its initial speed is 5.422 m/s.
Explanation:
Free falling body is the motion of a falling object under the influence of the Earth's gravity. Its motion is independent of its weight. The constant acceleration of a free falling body is called acceleration due to gravity, denoted by gg which is approximate to 9.8 m/s^{2}s
2
.
For the formula to be used in the problem, we use:
V_{1} ^{2}=V_{0} ^{2} +2gyV
1
2
=V
0
2
+2gy equation 1
y=V_{0} t+\frac{1}{2} gt^{2}y=V
0
t+
2
1
gt
2
equation 2
where
V_{1}V
1
is the final velocity, unit is in m/s
V_{0}V
0
is the initial velocity, unit is in m/s
tt is the time, unit is in seconds (s)
yy is the vertical distance, unit is in meters (m)
gg is the acceleration due to gravity, 9.8 m/s^{2}9.8m/s
2
For the given information
y=1.50my=1.50m height of the table
x=2.0mx=2.0m distance of the marble from the base of the table (when the
marble strikes the floor)
Solving the problem
1. To solve for the initial velocity, use equation 1 then substitute the given information.
V_{1} ^{2}=V_{0} ^{2} +2gyV
1
2
=V
0
2
+2gy
0=V_{0} ^{2} +2(-9.8m/s^{2} )(1.5m)0=V
0
2
+2(−9.8m/s
2
)(1.5m) gg is negative since the acceleration is
downward
V_{0} =\sqrt{(2(9.8)(1.5)}V
0
=
(2(9.8)(1.5)
V_{0} =5.422m/sV
0
=5.422m/s
2. To solve for the time, use equation 2 then substitute the value of
V_{0} =5.422 m/sV
0
=5.422m/s and the given, we get:
y=V_{0} t+\frac{1}{2} gt^{2}y=V
0
t+
2
1
gt
2
1.50m=(5.422m/s)t+\frac{1}{2} (9.8)t^{2}1.50m=(5.422m/s)t+
2
1
(9.8)t
2
Simplifying and arranging the equation
4.9t^{2} +5.422t-1.50=04.9t
2
+5.422t−1.50=0
Solving for tt using quadratic equation, x=\frac{-b+\sqrt{b^{2}-4ac } }{2a}x=
2a
−b+
b
2
−4ac
, we get:
t=\frac{-5.422+\sqrt{5.422^{2+4(4.9)(1.50)} } }{2(4.9)}t=
2(4.9)
−5.422+
5.422
2+4(4.9)(1.50)
t=0.553sec
