Mt. Asama, Japan, is an active volcano. In 2009, an eruption threw solid volcanic rocks that landed 1 km horizontally from the crater. If the volcanic rocks were launched at an angle of 40° with respect to the horizontal and landed 900 m below the crater, (a) what would be their initial velocity and (b) what is their time of flight?

a) 69.3 m/s

b) 18.84 s

Explanation:

Let the initial velocity = u

The vertical and horizontal components of the velocity is given by uᵧ and uₓ respectively

uᵧ = u sin 40° = 0.6428 u

uₓ = u cos 40° = 0.766 u

We're given that the horizontal distance travelled by the projectile rock (Range) = 1 km = 1000 m

The range of a projectile motion is given as

R = uₓt

where t = total time of flight

1000 = 0.766 ut

ut = 1305.5

The vertical distance travelled by the projectile rocks,

y = uᵧ t - (1/2)gt²

y = - 900 m (900 m below the crater's level)

-900 = 0.6428 ut - 4.9t²

Recall, ut = 1305.5

-900 = 0.6428(1305.5) - 4.9 t²

4.9t² = 839.1754 + 900

4.9t² = 1739.1754

t = 18.84 s

Recall again, ut = 1305.5

u = 1305.5/18.84 = 69.3 m/s

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2

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