Question

A dart leaves a toy dart gun with initial velocity of 7.76 m/s, regardless of the angle it is fired. What is the maximum horizontal distance that the dart can travel? Assume that the dart gun is fired at ground level.
A. 3.88m

B. More Information is needed

C. 6.14m

D. 12.28m

Answer 1

Answer:

C. 6.14 m

Explanation:

The maximum horizontal distance travelled by a projectile is given by:

$d=\frac{v^2}{g}sin (2\theta)$

where

v is the initial velocity

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta$ is the angle of launch

From the equation, we see that the maximum range is achieved when the projectile is fired at $\theta=45^{\circ}$.

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:

$d=\frac{7.76^2}{9.8}sin (2\cdot 45^{\circ}) = 6.14 m$