A volcanic eruption throws a boulder that lands 1.00 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 boulders initial velocity and (b) what is the time of flight?

a)  v₀ = 69.29 m / s , b)  t = 18.84 s

Explanation:

a) For this exercise we will use the projectile launch equations

          x = v₀ₓ t

          y = y₀ + t - ½ g t²

Let's fix our reference system on the volcano, so the horizontal distance x = 1 km = 1000 m and the vertical distance y = -900 m, the initial height of the crater is I = 0 m. Let's replace to find the speeds

         v_{oy} = v₀ sin θ

         v₀ₓ = v₀ cos θ

         y = v₀ sin θ (x / v₀ cos θ) - ½ g (x / v₀ cos θ)2

         y = x tan θ - ½ g x² / v₀² sec² θ

         ½ g x² sec² θ / v₀² = x tan θ - y

         v₀² = ½ g x² sec² θ / (x tan θ –y)

Let's calculate

           v₀² = ½ 9.8 1000² sec² 40 / (1000 tan 40 - (-900))

           v₀ = √ (8.35 10⁶ / 1,739 10³)

           v₀ = 69.29 m / s

    b) Flight time

           x = v₀ₓ t

           t = x / v₀ cos θ

           t = 1000 / 69.29 cos 40

           t = 18.84 s

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y