Answer:
-22.1
Step-by-step explanation:
1 / 4
Step 1. List horizontal (xxx) and vertical (yyy) variables
xxx-direction yyy-direction
t=\text?t=?t, equals, start text, question mark, end text t=\text?t=?t, equals, start text, question mark, end text
a_x=0a
x
=0a, start subscript, x, end subscript, equals, 0 a_y=-9.8\,\dfrac{\text m}{\text s^2}a
y
=−9.8
s
2
m
a, start subscript, y, end subscript, equals, minus, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction
\Delta x=255\,\text mΔx=255mdelta, x, equals, 255, start text, m, end text \Delta y=\text ?Δy=?delta, y, equals, start text, question mark, end text
v_x=v_{0x}v
x
=v
0x
v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript v_y=?v
y
=?v, start subscript, y, end subscript, equals, question mark
v_{0x}=120\,\dfrac{\text m}{\text s}v
0x
=120
s
m
v, start subscript, 0, x, end subscript, equals, 120, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction v_{0y}=0v
0y
=0v, start subscript, 0, y, end subscript, equals, 0
Note that there is no horizontal acceleration, so v_x=v_{0x}v
x
=v
0x
v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript. The time is the same for the xxx and yyy directions.
Also, the package has no initial vertical velocity.
Our yyy-direction variable list has too many unknowns to solve for \Delta yΔydelta, y directly. Since both the yyy- and xxx-directions have the same time ttt and horizontal acceleration is zero, we can solve for ttt from the xxx-direction motion by using equation:
\Delta x=v_xtΔx=v
x
tdelta, x, equals, v, start subscript, x, end subscript, t
Once we know ttt, we can solve for \Delta yΔydelta, y using the kinematic equation that does not include the unknown variable v_yv
y
v, start subscript, y, end subscript:
\Delta y=v_{0y}t+\dfrac {1}{2}a_yt^2Δy=v
0y
t+
2
1
a
y
t
2
delta, y, equals, v, start subscript, 0, y, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, start subscript, y, end subscript, t, squared
Hint #22 / 4
Step 2. Find ttt from horizontal variables
\begin{aligned}\Delta x&=v_xt \\\\ t&=\dfrac{\Delta x}{v_{0x}} \\\\ t&=\dfrac{255\,\text m}{120\dfrac{\text m}{\text s}} \\\\ &=2.125\,\text s \end{aligned}
Δx
t
t
=v
x
t
=
v
0x
Δx
=
120
s
m
255m
=2.125s
Hint #33 / 4
Step 3. Find \Delta yΔydelta, y using ttt
Using ttt to solve for \Delta yΔydelta, y gives:
\begin{aligned}\Delta y&=v_{0y}t+\dfrac{1}{2}a_yt^2 \\\\ &=\cancel{ (0 )t}+\dfrac{1}{2}\left (-9.8\dfrac{\text m}{\text s^2}\right )\left(2.125\,\text s\right)^2 \\\\ &=-22.1\,\text m \end{aligned}
Δy
=v
0y
t+
2
1
a
y
t
2
=
(0)t
+
2
1
(−9.8
s
2
m
)(2.125s)
2
=−22.1m
Hint #44 / 4
The correct answer is -22.1\,\text m−22.1mminus, 22, point, 1, start text, m, end text.