Answer:
The solutions are
=
0
t=0 and
=
14
4.9
≈
2.86
t=
4.9
14
≈2.86 seconds. Since time can't be negative, the ball was in the air for approximately 2.86 seconds.
Explanation:
To find out how long the ball was in the air, you can use the vertical motion equation:
ℎ
=
ℎ
0
+
0
−
1
2
2
h=h
0
+v
0y
t−
2
1
gt Where:
ℎ
h is the final height (1.6 m when caught),
ℎ
0
h
0
is the initial height (1.6 m when thrown),
0
v
0y
is the vertical component of the initial velocity (since it's thrown at an angle,
0
=
0
sin
(
)
v
0y
=v
0
sin(θ)),
g is the acceleration due to gravity (approximately
9.8
m/s
2
9.8m/s
2
),
t is the time in seconds.Given:
ℎ
=
1.6
m
h=1.6m,
ℎ
0
=
1.6
m
h
0
=1.6m,
0
=
28
m/s
v
0
=28m/s,
=
4
5
∘
θ=45
∘
,
=
9.8
m/s
2
g=9.8m/s
2First, find
0
v
0y
:
0
=
0
sin
(
)
v
0y
=v
0
sin(θ)
0
=
28
m/s
×
sin
(
4
5
∘
)
v
0y
=28m/s×sin(45
∘
)
0
=
28
m/s
×
2
2
v
0y
=28m/s×
2
2
0
=
14
m/s
v
0y
=14m/s
Now, plug in the values to solve for
t:
1.6
=
1.6
+
(
14
)
−
1
2
(
9.8
)
2
1.6=1.6+(14)t−
2
1
(9.8)t
2
This simplifies to:
0
=
14
−
4.9
2
0=14t−4.9t
2
Solve this quadratic equation for
t.
2