A stone is dropped from the top of a cliff. It hits the ground below after 3.25 s. How high is the cliff?

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Nov 11, 2015

Use the formula #d=1/2at^2#. This formula is valid for constant acceleration and no initial velocity. Use acceleration due to gravity#=g=9.8m/s^2#, and #t=3.25s#. Then #d=1/2(9.8m/s^2)(3.25s)^2=51.8m#.

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Nov 11, 2015

Answer:

#51.8m#

Explanation:

Lets list our variables first
#Deltay = ?m# This is what we are trying to find
#v_i=0m/s# The initial velocity is 0 because our object dropped from rest
#v_f=0# The final velocity is not relevant in this problem
#t = 3.25sec# We are given the time in seconds
#a = 9.80m/s^2# This is the downward acceleration due to gravity

Now we use our kinematic equation
#Deltay = v_i*t+(a*t^2)/2#
Plug in the variables

#Deltay = 0m/s*3.25sec+(9.80m/cancel(s^2)*3.25^2cancel(sec^2))/2#

#Deltay = 51.75m#

Because there are 3 significant figures, we round the answer to #51.8m#

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