91Ó°ÊÓ

$$\begin{gathered} 1)\mathrm{Angle}\mathrm{of}\mathrm{the}\mathrm{ejection},\mathrm{θ}=35° \\ 2)\mathrm{Vertical}\mathrm{distance},\mathrm{y}=3.30\mathrm{km}\mathrm{or}-3300\mathrm{m}\left(\mathrm{downard}\right) \\ 3)\mathrm{Horizontal}\mathrm{distance},\mathrm{x}=9.40\mathrm{km}\mathrm{or}9400\mathrm{m} \end{gathered}$$

The volcanic bomb has only gravitational force acting on it. Therefore, its acceleration is constant in the vertical direction and there is no acceleration in the horizontal direction. Hence, we can treat this motion as projectile motion.

Formulae:

The equation of distance in the projectile motion,$\mathrm{y}={\mathrm{³Ù²¹²Ôθ}}_0\mathrm{x}-\frac{{\mathrm{gx}}^2}{2{\left({\mathrm{V}}_0{\mathrm{³¦´Ç²õθ}}_0\right)}^2}$ …(¾±)

The time of flight of a body in projectile motion $\mathrm{T}=\frac{\mathrm{x}}{{\mathrm{V}}_0{\mathrm{³¦´Ç²õθ}}_0}$ …(¾±¾±)

By rearranging equation (i) of projectile motion, we get the value of the initial velocity of the volcanic bomb by substituting the given values as follows:

$$\begin{gathered} {\mathrm{V}}_0=\frac{9400\mathrm{m}}{\cos \left(35°\right)}\sqrt{\frac{9.8\mathrm{m}/{\mathrm{s}}^2}{2\left(9400\mathrm{m}×\tan (35°\right)-\left(-3300\mathrm{m})\right)}} \\ =255.52\mathrm{m}/\mathrm{s} \\ ≈260\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the initial velocity is 260 m/s.

Using the given values in equation (ii), we can get the value of the time of flight as follows:

$$\begin{gathered} \mathrm{T}=\frac{9400\mathrm{m}}{\left(255.52\mathrm{m}/\mathrm{s}\right)\cos \left(35°\right)} \\ =44.91\sec \\ ≈45\sec \end{gathered}$$

Hence, the value of time is 45 se c.

The air will oppose the motion of the volcanic bomb, hence we expect the initial speed required to reach the foot of the volcano will be increase.