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

2.55 m

Step-by-step 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=(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=((5)^2(sin 2\cdot 45^(\circ)))/(9.8)=2.55 m`