Question

A child bounces off of a trampoline with a velocity of 12.0 m/s at an angle of 37º and lands on another trampoline at the same height. What was the horizontal range of the child?

Answer 1

The child lands 14.12 meters apart from the first trampoline

Step-by-step explanation:

Projectile Motion

When an object is launched with angle respect to the horizontal it describes a curved path called a parabola. It reaches its maximum height and finally returns to the same high it was launched from. The horizontal distance covered in the movement is called the range and can be computed by

`\displaystyle x_(max)=(V_(o)^2\ sin2\theta)/(g)`

Where
`V_o` is the initial speed and
`\theta` the angle. The child bounces off of a trampoline with a velocity of
`V_o=12\ m/s` at an angle of
`\theta=37^o`. Knowing the child lands on another trampoline at the same height, the horizontal distance is

`\displaystyle x_(max)=(12^2\ sin[2(37^o)])/(9.8)=14.12\ m`

The child lands 14.12 meters apart from the first trampoline