91Ó°ÊÓ

$\mathrm{θ}=53.0°$

time required to reach ground$=5.00s$

altitude of the plane$=730m$

As the projectile is released from the plane, so projectile will have the same velocity as that of the plane.

Using the kinematic equations and the given datathe initial velocity of the projectile can be computed. Once, the initial velocity of the projectile is found.the required answers can be found.

Formulae:

The vertical distance in kinematic equation can be written as,

$d_y=v_{0}t+\frac{1}{2}×a{t}^2$ (i)

Final velocity is

$v_f^2=v_0^2+2a{b}_y$

(ii)

$v_f=v_0+at$

(iii)

$v=\frac{d}{t}$

(iv)

Assuming downward as positive, Equation (i) can be written as,

$$\begin{gathered} 730={\mathrm{v}}_0×\cos \left(53\right)+\frac{1}{2}×9.8×5^2 \\ 730=3{\mathrm{v}}_0+122.5 \\ 3v_0=730-122.5 \\ 3{\mathrm{v}}_0=607.5 \\ {\mathrm{v}}_0=\frac{607.5}{3} \\ {\mathrm{v}}_0=202\mathrm{m}/s \end{gathered}$$

The velocity in x direction is given by,

$$\begin{gathered} v_{0x}=202×\sin \left(53\right) \\ {v}_{0x}=161.321m/s \end{gathered}$$

Using equation (iv), the distance in x direction is

$$\begin{gathered} d_x=v_{0x}×t \\ =161.32×5 \\ {d}_x=806m \end{gathered}$$

As there is no acceleration in the y direction, the x component for the projectile and plane will be same,

$$\begin{gathered} v_{0x}=202×\sin \left(53\right) \\ {v}_{0x}=161.32 \\ =161m/s \end{gathered}$$

Now, with equation (iii) the final velocity in y direction is given by,

$$\begin{gathered} v_y=v_{0y}+at \\ =-202×\cos \left(53\right)-9.8×5 \\ =-121.57-49 \end{gathered}$$

Thus,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{y}}=-170.57 \\ ≈-171\mathrm{m}/\mathrm{s} \end{gathered}$$