Answer:−47.0
Step-by-step explanation: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=12\,\text mΔx=12mdelta, x, equals, 12, 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=\text ?v
y
=?v, start subscript, y, end subscript, equals, start text, question mark, end text
v_{0x}=2.5\,\dfrac{\text m}{\text s}v
0x
=2.5
s
m
v, start subscript, 0, x, end subscript, equals, 2, point, 5, 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, and the time is the same for the xxx- and yyy-directions.
Also, the pumpkin has no initial vertical velocity.
Our yyy-direction variable list has too many unknowns to solve for v_yv
y
v, start subscript, y, end subscript 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 v_yv
y
v, start subscript, y, end subscript using the kinematic equation that does not include the unknown variable \Delta yΔydelta, y:
v_y=v_{0y}+a_ytv
y
=v
0y
+a
y
tv, start subscript, y, end subscript, equals, v, start subscript, 0, y, end subscript, plus, a, start subscript, y, end subscript, t
Hint #22 / 4
Step 2. Find ttt from horizontal variables
\begin{aligned}\Delta x&=v_{0x}t \\\\ t&=\dfrac{\Delta x}{v_{0x}} \\\\ &=\dfrac{12\,\text m}{2.5\,\dfrac{\text m}{\text s}} \\\\ &=4.8\,\text s \end{aligned}
Δx
t
=v
0x
t
=
v
0x
Δx
=
2.5
s
m
12m
=4.8s
Hint #33 / 4
Step 3. Find v_yv
y
v, start subscript, y, end subscript using ttt
Using ttt to solve for v_yv
y
v, start subscript, y, end subscript gives:
\begin{aligned}v_y&=v_{0y}+a_yt \\\\ &=\cancel{0\,\dfrac{\text m}{\text s}}+\left(-9.8\,\dfrac{\text m}{\text s}\right)(4.8\,\text s) \\\\ &=-47.0\,\dfrac{\text m}{\text s} \end{aligned}
v
y
=v
0y
+a
y
t
=
0
s
m
+(−9.8
s
m
)(4.8s)
=−47.0
s
m
is the common difference between terms. You can solve for 
in terms of any of the previous terms of the sequence:
and the 11th term is 
, you have the following system of equations.
and 
.