Answer:
The angle of projection is 12.26⁰.
Explanation:
Given;
initial position of the dart, h₀ = 1.50 m
height above the ground reached by the dart, h₁ = 1.73 m
maximum height reached by the dart, Hm = h₁ - h₀ = 1.73 m - 1.50 m= 0.23 m
velocity of the dart, u = 10 m/s
The maximum height reached by the projectile is calculated as;
[tex]H_m = \frac{u^2sin^2 \theta}{2g}[/tex]
where;
θ is angle of projection
g is acceleration due to gravity = 9.8 m/s²
[tex]H_m = \frac{u^2sin^2 \theta}{2g}\\\\sin^2 \theta = \frac{H_m \ \times \ 2g}{u^2} \\\\sin^2 \theta = \frac{0.23 \ \times \ 2(9.8)}{10^2} \\\\sin ^2\theta =0.04508\\\\sin \theta = \sqrt{0.04508} \\\\sin \theta = 0.2123\\\\\theta = sin^{-1}(0.2123)\\\\\theta = 12.26^0[/tex]
Therefore, the angle of projection is 12.26⁰.
