a plane drops a package of supplies to a party of explorers. if the plane is travelling horizontally at 40m/s and is 100m above the ground where does the package strike the ground relative to the point where it released?? help plzzzz

Question

aaronq · Answer

Once released, the package is in free-fall and its motion can be analyzed by separating the  horizontal and vertical components. 

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In the question, you're told that the plane is moving at $\sf v_x=100 ~m/s$, this is the horizontal component. The object is also accelerating in the vertical direction at $\sf a_y=-9.8 ~m/s^2$ (Note the negative sign because its in the negative y direction).

We don't how long that takes, but we can use the distance and the acceleration in the y direction.

$\sf \Delta y=v_{y}\Delta t +\dfrac{1}{2}a_y(\Delta t)^2$\)

$\sf -100~m=(0)\Delta t +\dfrac{1}{2}(-9.8~m/s^2)(\Delta t)^2$

$\sf \Delta t=4.52~s$

We can use the time for it to hit the ground with $\sf v_x$ to find the distance it travelled.

$\sf x_f=x_i+v_x\Delta t +\dfrac{1}{2}a_x(\Delta t)^2$

$\sf x_f=0+(40~m/s)(4.52~s)+\dfrac{1}{2}(0)(4.52~s)^2=180.702~\approx~181~m$

The package travelled 181 meters in the horizontal direction