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?

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barnuts barnuts · Answer 1 · 2017-10-17T13:38:27+02:00

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

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batolisis batolisis · Answer 2 · 2021-10-04T13:52:42+02:00

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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