Answer:
θ = 18.3°
Explanation:
The angle can be found by using the formula for the range of projectile motion:
[tex]R = \frac{u^2Sin2\theta}{g}[/tex]
where,
R = Horizontal Range = 8.75 m
u = launch speed = 12 m/s
g = acceleration due to gravity = 9.81 m/s²
θ = launch angle = ?
Therefore,
[tex]8.75\ m = \frac{(12\ m/s)^2\ Sin2\theta}{9.81\ m/s^2} \\\\Sin2\theta = \frac{(8.75\ m)(9.81\ m/s^2)}{(12\ m/s)^2} \\\\2\theta = Sin^{-1}(0.6)\\\theta = \frac{36.6^o}{2}[/tex]
θ = 18.3°
Since this angle is less than 45°. Therefore, it is the smaller of the two possible angles an dit is the correct answer:
θ = 18.3°
