Question

An airplane flying horizontally at a velocity of 138 m/s and at an altitude of 1500 meters when one of its wheels falls off.What horizontal distance (in meters) will the wheel travel before it strikes the ground?

Answer 1

2415 meters

Explanation:

Given:

Initial horizontal velocity (ux) = 138 m/s

Initial vertical velocity (uy) = 0 m/s

Height (h) = 1500 meters

The first thing is to calculate the time it takes for the airplane to reach the ground, just like this:

`\begin{gathered} h=u_yt+(1)/(2)at^2 \\ \\ \text{ we replacing} \\ \\ 1500=0\cdot t+(1)/(2)(9.8)t^2 \\ \\ 4.9t^2=1500 \\ \\ t^2=(1500)/(4.9) \\ \\ t=\sqrt{(1500)/(4.9)} \\ \\ t=17.5\text{ sec} \end{gathered}`

Therefore, the horizontal distance would be:

`\begin{gathered} x=u_x\cdot t \\ \\ \text{ we replacing} \\ \\ x=138\cdot17.5 \\ \\ x=2415\text{ m} \end{gathered}`

Therefore, the horizontal distance is 2415 meters