Amodel 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?

First calculate for the time it takes for the model rocket to reach the land. Using the formula:

h = vi t + 0.5 g t^2

where h is height = 100 m, vi is initial vertical velocity = 0 since it simply dropped, t is time = ?, g = 9.8 m/s^2

100 = 0.5 (9.8) t^2

t = 4.52 seconds

So the total horizontal distance taken is:

d = 50 m/s * 4.52 s

d = 225.90 m

The distance from the edge of the cliff where the model rocket will land is ; 225.90 m

Using the given data :

Velocity of rocket = 50.0 m/s

Depth of canyon ( H ) = 100 m deep

Initial velocity ( vi ) = 0

To determine how far from the edge of the cliff does the model rocket land

H = vi( t ) + 0.5gt²

100 = 0.5*(9.8)*t²

∴ t = 4.52  secs

Velocity = horizontal distance / time

∴ distance ( horizontal distance ) = velocity * time

   distance = 50 m/s * 4.52 secs = 225.90 m

Hence we can conclude that The distance from the edge of the cliff where the model rocket will land is ; 225.90 m .

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unbalanced force

explanation:

increase masses, decrease separation