Question

A) The volcano to the right shoots out a chunk of hot magma at an angle of theta zero equals 35° Above the horizontal from point A as shown the hot magma reaches point B in a time of t =45 seconds if H=3.30 km find the initial speed of the hot magma. B) what horizontal distance D in the image did the hot magma travel?

Answer 1

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,

`T=(2u\sin \theta_D)/(g)`

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,

`\begin{gathered} 45=(2* u*\sin 35\degree)/(9.8) \\ \Rightarrow u=(45*9.8)/(2*\sin 35\degree) \\ =384.43\text{ m/s} \end{gathered}`

The range of the projectile or the horizontal distance traveled by the magma is given by,

`D=(u^2\sin 2\theta_D)/(g)`

On substituting the known values,

`\begin{gathered} D=(384.43^2*\sin (2*35\degree))/(9.8) \\ =14170.8\text{ m} \\ =14.17km \end{gathered}`

Thus the magma travels for a horizontal distance of 14.17 km