Question

A motorcyclist is trying to leap across a canyon by driving horizontally off the higher side of the canyon at a speed of 78.0 m/s. If air resistance is neglected and the difference in height on two sides of the canyon is 252 m, what is the speed of the motorcycle as it reaches the lower side of the canyon?

Answer 1

Answer:

v = (78.0 i ^ - 70.27 j ^) m/s,      v = 105 m / s ,    θ = 318º

Explanation:

We have a projectile launch problem, let's start by calculating the time it takes to get through the canyon

        y = $v_{oy}$ t - ½ gt2

As the motorcyclist comes out horizontally, the speed he has is the horizontal speed (vox) and the initial vertical speed is zero (I go = 0)

       y = 0 - ½ g t2

       t = √ 2y / g

       t = √ (2 252 /9.8)

       t = 7.17 s

Let's calculate the vertical speed for this time

        $v_{y}$  =  $v_{oy}$  - gt

        $v_{y}$  = 0 - gt

       $v_{oy}$  = - 9.8 7.17

       $v_{oy}$  = - 70.27 m / s

We can give the result in two ways

First:

    v = (78.0 i ^ - 70.27 j ^) m / s

Second:

using the Pythagorean theorem and trigonometry

   v² = vₓ² +  $v_{oy}$ ²

   v = √ [(78.0)² + (-70.42)²] = √ (11042.98)

   v = 105 m / s

   tan θ₁ =  $v_{y}$ y / vₓ

   tan θ₁ = -70.42 / 78.0

   θ₁ = 42º

If we measure this angle from the positive direction of the x-axis counterclockwise

    θ = 360 - θ₁

    θ = 360 - 42

    θ = 318º