Question

An explosive projectile is launched straight upward to a maximum height h. At its peak, it explodes, scattering particles in all directions. Assume each particle has the same initial speed v after the explosion. What is the smallest angle that the final velocity of a particle will make with the horizontal upon hitting the ground

Answer 1

Answer:

θ = tan⁻¹ ($\frac{19.6 \ h}{v}$)

Explanation:

This problem must be solved using projectile launch ratios. Let's analyze the situation, the projectile explodes at the highest point, therefore we fear the height (i = h), the speed at this point is the same, but the direction changes, we are asked to find the smallest angle of the speed in the point of arrival with respect to the x-axis.

The speed at the arrival point (y = 0)

           v² = vₓ² + v_y²

Let's see how this angle changes, for two extreme values:

* The particle that falls from the point of explosion, in this case the speed is vertical

         v = v_y

the angle with the horizontal is 90º

* The particle leaves horizontally from the point of the explosion, the initial velocity is horizontal

         vₓ = v

the final velocity for y = 0

         v_f = vₓ² + v_y²

therefore the angle has a value greater than zero and less than 90º

As they ask for the smallest angle, we can see that we must solve the last case

the output velocity is horizontal vₓ = v

Let's find the velocity when it hits the ground y = 0, with y₀ = h

            $v_{y}^2$ = $v_{oy}^2$ - 2 g (y-y₀)

            v_{y}^2 = - 2g (0- y₀)

let's calculate

           v_{y}^2 = 2 9.8 h

         

we use trigonometry to find the angle

        tan θ = $\frac{v_y}{v_x}$

        θ = tan⁻¹ ($\frac{v_y}{v_x}$)

let's calculate

         θ = tan⁻¹ ($\frac{19.6 \ h}{v}$)