Answer:
6.2 seconds
Step-by-step explanation:
To find out how long it takes for the ball to reach the ground when thrown straight up from the top of a 64 m tall building with an initial velocity of 20 m/s, you can use the following kinematic equation:
ℎ
=
�
�
+
1
2
�
�
2
h=ut+
2
1
gt
2
Where:
ℎ
h is the height (negative in this case because it's falling downward, so it's -64 m),
�
u is the initial velocity (20 m/s),
�
g is the acceleration due to gravity (-9.8 m/s², taken as negative because it acts downward), and
�
t is the time you want to find.
Plug in the values:
−
64
=
(
20
)
�
+
1
2
(
−
9.8
)
�
2
−64=(20)t+
2
1
(−9.8)t
2
Now, rearrange and solve for
�
t:
−
64
=
20
�
−
4.9
�
2
−64=20t−4.9t
2
This is a quadratic equation. Rearrange it to make it equal to zero:
4.9
�
2
−
20
�
−
64
=
0
4.9t
2
−20t−64=0
Now, you can solve this quadratic equation for
�
t. You can use the quadratic formula:
�
=
−
�
±
�
2
−
4
�
�
2
�
t=
2a
−b±
b
2
−4ac
In this equation,
�
=
4.9
a=4.9,
�
=
−
20
b=−20, and
�
=
−
64
c=−64. Plug these values into the formula:
�
=
−
(
−
20
)
±
(
−
20
)
2
−
4
(
4.9
)
(
−
64
)
2
(
4.9
)
t=
2(4.9)
−(−20)±
(−20)
2
−4(4.9)(−64)
Now, calculate it:
�
=
20
±
400
+
1254.4
9.8
t=
9.8
20±
400+1254.4
�
=
20
±
1654.4
9.8
t=
9.8
20±
1654.4
Now, calculate the two possible values of
�
t by taking both the positive and negative square roots:
�
1
=
20
+
1654.4
9.8
t
1
=
9.8
20+
1654.4
�
2
=
20
−
1654.4
9.8
t
2
=
9.8
20−
1654.4
Calculating
�
1
t
1
:
�
1
≈
20
+
40.68
9.8
≈
60.68
9.8
≈
6.2
seconds
t
1
≈
9.8
20+40.68
≈
9.8
60.68
≈6.2 seconds
Calculating
�
2
t
2
:
�
2
≈
20
−
40.68
9.8
≈
−
20.68
9.8
≈
−
2.1
seconds
t
2
≈
9.8
20−40.68
≈
9.8
−20.68
≈−2.1 seconds
The negative value of
�
t doesn't make physical sense in this context, so we consider the positive value.
It takes approximately 6.2 seconds for the ball to reach the ground.