Respuesta :
Answer:
θ = tan⁻¹ ([tex]\frac{19.6 \ h}{v}[/tex])
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
[tex]v_{y}^2[/tex] = [tex]v_{oy}^2[/tex] - 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 = [tex]\frac{v_y}{v_x}[/tex]
θ = tan⁻¹ ([tex]\frac{v_y}{v_x}[/tex])
let's calculate
θ = tan⁻¹ ([tex]\frac{19.6 \ h}{v}[/tex])
