Mt. asama, japan, is an active volcano complex. in 2009, an eruption threw solid volcanic rocks that landed far from the crater. suppose that one such rock was launched at an angle of θ = 58.7 degrees above horizontal, and landed a horizontal distance d = 1200 m from the crater, and a vertical distance h = 780 m below the crater.

write and expression for v0, the initial speed of the rock in terms of θ, d, and h.

The expression of Vo in terms of ∅ , d and h is ; V₀ =

( i.e. Vo = 97.41 m/s )

Given data :

∅  or  β  = 58.7°

Horizontal distance ( d ) = 1200 m  

Vertical height ( h ) = 780 m

Note :  ∅ = β ,  y = h   in the expression of Vo

First step : Expressing/calculating  the Vertical distance ( y )

y = u sin∅t -   ----- ( 1 )

next ; Expressing horizontal distance

x = u cos∅t  + 0 ----- ( 2 )

∴ t = x / u cos∅  ----- ( 3 )

Insert equation ( 3 ) into equation ( 1 )

y = u sinβ ( x / u cosβ )  -  

  ∴ U ( V₀ ) =  =  97.41 m/s

where ; y = -780 m,  x = 1200 m.

Hence we can conclude that The expression of Vo in terms of ∅ , d and h is ; V₀ =    ( i.e. Vo = 97.41 m/s )

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u=97.41m/s

Explanation:

Given

inclination

Horizontal distance travel by Particle

Vertical height

Let u be the initial velocity

calculating vertical distance

-------1

Calculating horizontal distance

put value of t in equation 1

at

the answer is 12 as i may remember

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