Question

The pilot of an airplane traveling 160km/h wants to drop supplies to flood victims isolated on a patch of land 160 m below . The supplies should be dropped how many seconds before the plane is directly overhead ?

Answer 1

We are given that an airplane travelling at 160 km/h and 160 meters high will drop supplies. The problem can be exemplified in the following diagram:

Due to inertia, the movement of the supplies will be that of a parabolic motion. therefore, we can use the following equation of motion:

$y=y_0+v_yt-\frac{1}{2}gt^2$

Where:

$\begin{gathered} y=\text{ height} \\ v_{0y}=\text{ initial vertial velocity} \\ g=\text{ acceleration of gravity} \\ t=\text{ time} \end{gathered}$

Since the plane travels horizontally, this means that the vertical velocity of the object is zero, therefore, we have:

$y=y_0-\frac{1}{2}gt^2$

The value of the height "y" is zero since we want to determine the time when the object hits the ground, therefore we have:

$0=y_0-\frac{1}{2}gt^2$

Now we solve for the time "t" first by subtracting the initial height from both sides:

$-y_0=-\frac{1}{2}gt^2$

Now we multiply both sides by -2:

$2y_0=gt^2$

Now we divide both sides by "g":

$\frac{2y_0}{g}=t^2$

Now we take the square root to both sides:

$\sqrt[]{\frac{2y_0}{g}}=t$

Replacing the given values we get:

$\sqrt[]{\frac{2(160m)}{9.8\frac{m}{s^2}}}=t$

Solving the operations we get:

$5.71s=t$

Therefore, the supplies must be dropped 5.71s before the plane is directly overhead.