Respuesta :
Theprojectile launch allows to find the correct answer for the question about the direction of the ball when it hits the ground is:
D) θ = 37º measured clockwise from the x-axis.
projectile launch is an application of kinematics where there is no acceleration on the horizontal axis and the acceleration on the vertical axis is the gravity acceleration.
The horizontal distance traveled until it reaches the ground is
x = v₀ₓ t
The height of the bullet is
y = y₀ + [tex]v_o_y[/tex] t - ½ g t²
The initial vertical velocity is zero and when it reaches the ground its height is zero.
0 = y₀ - ½ g t²
t = [tex]\sqrt{\frac{2y_o}{g} }[/tex]
We substitute
[tex]x = v_{ox} \ \sqrt{\frac{2y_o}{g} }[/tex]
They indicate that the distance is:
x = [tex]\frac{8}{3} \ y_o[/tex]
We substitute
[tex]\frac{8}{3} y_o = v_{ox} \sqrt{\frac{2y_o}{g} }[/tex]
[tex]v_{ox} = \frac{8}{3} \sqrt{\frac{y_o g}{2} }[/tex]
The vertical speed when reaching the ground is:
[tex]v_y = v_{oy} - g t \\v_y = - g t \\v_y = - g \sqrt{\frac{2y_o}{g} }[/tex]
Let's use trigonometry for the direction at the point of hitting the ground.
tan θ = [tex]\frac{v_y}{v_x}[/tex]
θ = tan⁻¹ ( [tex]- \frac{3}{8} \ \frac{\sqrt{2y_o/g}}{\sqrt{y_o g /2} }[/tex] )
θ = tan⁻¹ ( [tex]tan ^{-1} (- \frac{3}{8} \sqrt{4} )= tan^{-1 } \ (- \frac{3}{4} )[/tex]
θ = -36.9º
The negative sign indicates that the angle is clockwise from the x-axis.
In conclusion, using the projectile launch we can find the correct answer for the question about the direction of the bullet when it hits the ground is:
D) θ = 37º measured clockwise from the x-axis.
Learn more about projectile launch here: brainly.com/question/16848061
