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-(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-(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-(1)/(2)gt^2`

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

`-y_0=-(1)/(2)gt^2`

Now we multiply both sides by -2:

`2y_0=gt^2`

Now we divide both sides by "g":

`(2y_0)/(g)=t^2`

Now we take the square root to both sides:

`\sqrt[]{(2y_0)/(g)}=t`

Replacing the given values we get:

`\sqrt[]{(2(160m))/(9.8(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.