Question

A model rocket flies horizontally off the edge of a cliff at a velocity of 50.0 m/s. If the canyon below is 100.0 m deep, how far from the edge of the cliff does the model rocket land?

My teacher provided the equations:
x= vt and y = (Vi * t)1/2* a * t^2

Answer 1

Yes! The second one is one of the constant acceleration equations. We know that our X and Y values are independent of each other, which is why Y doesn't matter here. We only care about the X position value since X and Y are independent! (This is an extremely important rule, trust me)

Therefore, we don't need to calculate a Y velocity or anything like that. We only need Y because we need to find time t. We can use Y to calculate time because although X and Y are independent of each other, they do share the same time traveled!

We know that
`V_(0,y)` = 0, since the car wasn't moving vertically initially. We also know Y initial is 100 and Y final is 0 when it hits the ground! The last thing we know is that the acceleration of any item in free fall is (usually) -9.81. Therefore, using the original Y equation that your teacher gave, we can plug things in and solve for time!

`y_f=y_i+v_it+(1)/(2)at^2\\0=100+(0)t+(1)/(2)(-9.81)t^2\\-100=-4.905t^2\\t^2=20.39\\t=4.52`t = 4.52s

(Your teacher simplified the Y equation a little, so it may look slightly different from mine. I started at the original Y equation)

Now we have time! So now we can use the other equation your teacher provided. Again, remember, even though the car is now traveling through the air vertically, that does not affect its x velocity! The car is still traveling at 50m/s!

Therefore, we can just plug and chug!:

x = vt

x = (50)(4.52)

x = 226m

I hope this helps!