Answer:
C. 6.14 m
Explanation:
The maximum horizontal distance travelled by a projectile is given by:
[tex]d=\frac{v^2}{g}sin (2\theta)[/tex]
where
v is the initial velocity
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]\theta[/tex] is the angle of launch
From the equation, we see that the maximum range is achieved when the projectile is fired at [tex]\theta=45^{\circ}[/tex].
The dart in the problem has an initial velocity of
v = 7.76 m/s
Substituting into the formula, we find the maximum horizontal distance:
[tex]d=\frac{7.76^2}{9.8}sin (2\cdot 45^{\circ}) = 6.14 m[/tex]
