Given,
The angle of projection of the magma chunk, θ_D=35°
The time of flight, t=45 s
The height from which the magma chunk was projected, H=3.30 km=3300 m
The time of flight is given by,
[tex]T=\frac{2u\sin \theta_D}{g}[/tex]
Where u is the initial velocity with which the Magna chunk was projected and g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} 45=\frac{2\times u\times\sin 35\degree}{9.8} \\ \Rightarrow u=\frac{45\times9.8}{2\times\sin 35\degree} \\ =384.43\text{ m/s} \end{gathered}[/tex]
The range of the projectile or the horizontal distance traveled by the magma is given by,
[tex]D=\frac{u^2\sin 2\theta_D}{g}[/tex]
On substituting the known values,
[tex]\begin{gathered} D=\frac{384.43^2\times\sin (2\times35\degree)}{9.8} \\ =14170.8\text{ m} \\ =14.17km \end{gathered}[/tex]
Thus the magma travels for a horizontal distance of 14.17 km
