A plane drops a package for delivery. The plane is flying horizontally at a speed of 120\,\dfrac{\text m}{\text s}120 s m 120, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction, and the package travels 255\,\text m255m255, start text, m, end text horizontally during the drop. We can ignore air resistance.

-22.1

Explanation:

1 / 4

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=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 subscriptv_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 fractionv_{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.

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge