Given,
The angle of projection, θ_D=35.0°
The time of flight, T=45.0 s
The height, h=3.30 km=3300 m
A.
The time of flight of a projectile is given by,
[tex]T=\frac{2u\sin \theta}{g}[/tex]
Where u is the initial velocity of the magma chunk and g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} 45.0=\frac{2\times u\times\sin 35.0\degree}{9.8} \\ \Rightarrow u=\frac{45.0\times9.8}{2\times\sin 35.0\degree} \\ =384.43\text{ m/s} \end{gathered}[/tex]
Thus the initial speed of the magma chunk is 384.43 m/s
B. The horizontal distance of the flight of the magma chunk is given by,
[tex]\begin{gathered} R=u_xT \\ =u\cos \theta\times T \end{gathered}[/tex]
Where u_x is the horizontal component of the initial velocity.
On substituting the known values,
[tex]\begin{gathered} R=384.43\times\cos 35.0\degree\times45.0 \\ =14170.8\text{ m} \\ =14.17km \end{gathered}[/tex]
Thus the horizontal distance of the magma is 14.17 km
