Volcanoes on the Earth eject rocks at speeds of up to 50 m/s. Consider a 1500 m high volcano which ejects rocks in all directions. What is the maximum horizontal distance at sea level reached by the rocks?

a) V=gt and H=0.5 g t^2

h= 0.5 g (V/g)^2

H= 1000+ h=

H= 1000 + (V/)^2 /(2 g)

H= 1620 m

The maximum horizontal distance at sea level reached by the rocks depends on the horizontal component and it was not specified. Coriolis effect is negligible.

A fanout can be estimated if the geometry of the vent is known and the size of the caldera. Typical fanout can be assumed at 10 to 15 degrees. Then the we have to resolve the initial velocity into vertical and horizontal components.

Vv= V cos(15) using max angle

Vh= Vsin(15)

Then

H= 1000 + (Vcos(15))^2 /(2 g)

H= 1000 + 575= 1575m

S=Vsin(15) t where

t= t1 + t2

t1- time to travel to the upper most point h=(Vcos(15))^2 /(2 g)

t2 - time to fall height H

In general t=sqrt(2h/g)

t1= sqrt(2h/g)= sqrt(2 x 575 / 9.81)=11 s

t2= sqrt(2H/g) = sqrt(2 x 1575 /9.81)=18s

t= 29 s

S= V sin(15) t=

S= 110 sin(15) x 29= 830m max for a 15 degree fanout.

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.

reserve is the answer