A model rocket flies horizontally off the edge of the 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?

1 Answer
Nathan L.
May 31, 2017

#247# #"m"#

Explanation:

We're asked to find the distance #r# the object is from the edge of the cliff (where it was launched from) after being launched #50.0"m"/"s"# straight horizontally.

Since the object was launched horizontally, its initial velocity has no #y#-component (its #y#-component is #0#), so

  • #v_(0x) = v_0 = 50.0"m"/"s"#

  • #v_(0y) = 0#

We first need to find the time #t# at which the object hits the bottom of the #100.0"m"#-deep canyon. We can use the equation

#Deltay = v_(0y)t -1/2g t^2#

to find this. The change in #y#-position #Deltay# is #-100.0"m"# (it went downward), and since the initial #y#-velocity is #0#, the equation becomes

#-100.0"m" = -1/2g t^2#

which is the same as

#100.0"m" = 1/2g t^2#

We know that #g# is #9.80"m"/("s"^2)#, so

#100.0"m" = (4.90"m"/("s"^2))t^2#

#t = sqrt((100.0cancel("m"))/(4.90cancel("m")/("s"^2))) = color(red)(4.52"s"#

Therefore, the rocket hits the canyon's bottom after #4.52# seconds.

Using this and the formula

#x = v_xt#

we can find how far horizontally the rocket landed from the cliff:

#x = (50.0"m"/"s")(4.52"s") = color(green)(226"m"#

And the vertical distance is simply the height, #100.0"m"#

Finally, we can find the distance #r# the rocket lands from its launch point at the edge of the cliff using the formula

#r = sqrt(x^2 + y^2#

So,

#r = sqrt((226"m")^2 + (100.0"m")^2) = color(blue)(247"m"#

Thus, the rocket lands a distance of #247"m"# from the edge of the cliff.

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