Answer:
θ = 39.8°
Explanation:
Her, we will use the trigonometric ratio of the tangent to find the launch angle of the projectile from the vertical:
[tex]tan\ \theta = \frac{v_y}{v_x}[/tex]
where,
θ = launch angle from the vertical = ?
vy = vertical component of velocity = 5 m/s
vx = horizontal component of velocity = 6 m/s
Therefore,
[tex]tan\ \theta = \frac{5\ m/s}{6\ m/s}\\\\\theta = tan^{-1}(0.8333)[/tex]
θ = 39.8°
