91Ó°ÊÓ

Airplane speed is 350 km/h ( horizontally) .

It deals with the kinematic equations of motion in which the motion can be described with constant acceleration. Also, this problem is based on the concept of projectile motion. When a particle is thrown near the earth’s surface, it travels along a curved path under constant acceleration directed towards the center of the earth surface. This is a projectile motion problem, which contains two independent motions.

Formulae:

Kinematic equations:

$$\begin{gathered} v_{fy}=v_{0y}+at \\ ∆y=y_{0y}t+\left(\frac{1}{2}{a}_y{t}^2\right) \end{gathered}$$

Airplane is moving horizontally; hence initially bundle has speed only along horizontal direction.

Hence,

$v_{0y}=0$

And horizontal component will be same as airplanes speed as it was in the plane moving along with it.

Hence,

$v_{0x}=350\mathrm{km}/\mathrm{h}$

It is known that, horizontal motion would occur with constant speed as there is no acceleration in the horizontal direction.

Hence,

$$\begin{gathered} v_{fx}=v_{0x} \\ =350\mathrm{km}/\mathrm{h} \end{gathered}$$

$$\begin{gathered} ∆y=v_{0y}t+\left(1/2a_y{t}^2\right) \\ \mathrm{As},V_{iy}=0 \\ ∆y=\left(\frac{1}{2}{a}_y{t}^2\right) \end{gathered}$$

From this formula we can say that, time of fall depends on vertical height only, hence it would be same for any speed.