Answer:
2.55 m
Explanation:
The motion of the dancer is the motion of a projectile, which consists of 2 independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- A uniformly accelerated motion (constant acceleration) along the vertical direction
The horizontal range of a projectile can be found by using the equations of motions along the two directions, and it is given by:
[tex]d=\frac{v^2 sin(2\theta)}{g}[/tex]
where
v is the intial velocity
[tex]\theta[/tex] is the angle of projection
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
For the dancer in this problem, we have:
v = 5 m/s
[tex]\theta=45^{\circ}[/tex]
Therefore, the horizontal range is:
[tex]d=\frac{(5)^2(sin 2\cdot 45^{\circ})}{9.8}=2.55 m[/tex]