A stone is thrown with a horizontal velocity of 30.0 m/s from the top of a cliff 80.0 m high. How long will it take to strike the level ground at the base of the cliff, and how far from the foot of the cliff will it strike?

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Answer 1 · 2015-12-03T23:14:57

(a)

#4.04"s"#

(b)

#121.2"m"#

(a)

Taking the vertical component of motion we can assume that the initial velocity is zero so:

#s=1/2"g"t^2#

#:.t=sqrt((2s)/(g))#

#t=sqrt((2xx80)/(9.8))#

#t=4.04"s"#

(b)

The horizontal component of velocity is constant so:

Distance = speed x time

#:.d=30xx4.04=121.2"m"#