Question

The 10 kg dancer leaps into the air with an initial velocity of 5 m/s at angle of 45° from the floor. How far will she travel in the air horizontally before she lands on the ground again?

Answer 1

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:

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

where

v is the intial velocity

$\theta$ is the angle of projection

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

For the dancer in this problem, we have:

v = 5 m/s

$\theta=45^{\circ}$

Therefore, the horizontal range is:

$d=\frac{(5)^2(sin 2\cdot 45^{\circ})}{9.8}=2.55 m$