Answer:
The range is 35.35 m
Explanation:
Projectile Motion
It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.
Being vo the initial speed of the object, θ the initial launch angle, and [tex]g=9.8m/s^2[/tex] the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:
[tex]\displaystyle d={\frac {v_o^{2}\sin(2\theta )}{g}}[/tex]
The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:
[tex]\displaystyle d={\frac {20^{2}\sin(2\cdot 30^\circ )}{9.8}}[/tex]
[tex]\displaystyle d={\frac {400\sin(60^\circ )}{9.8}}[/tex]
[tex]d=35.35\ m[/tex]
The range is 35.35 m
