Answer:
A. 1.78 m
Explanation:
Given;
initial velocity of the, u = 12 m/s
angle of projection, θ = 30°
the range of the projectile, R = 7.5 m
To obtain the height through which the projectile lands, we must calculate the maximum height reached by the projectile.
[tex]H = \frac{u^2\ sin^2\theta}{2g} \\\\H = \frac{(12)^2\times (sin(30))^2}{2 \times 9.8} \\\\H = \frac{144 \times (0.5)^2}{19.6} \\\\H = 1.84 \ m[/tex]
Therefore, the height is most nearly 1.78 m
