A pumpkin is thrown horizontally off of a building at a speed of 2.5\,\dfrac{\text m}{\text s}2.5 s m 2, point, 5, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction and travels a horizontal distance of 12\,\text m12m12, start text, m, end text before hitting the ground. We can ignore air resistance.

−47.0

Step-by-step explanation:Step 1. List horizontal (xxx) and vertical (yyy) variables

xxx-directionyyy-direction

t=\text?t=?t, equals, start text, question mark, end textt=\text?t=?t, equals, start text, question mark, end text

a_x=0a

x

=0a, start subscript, x, end subscript, equals, 0a_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 subscriptv_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 fractionv_{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

step-by-step explanation:

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.