An object is dropped from the top of a building that is 64 feet tall. 

Use s(t) = -16t^2 + v0t+s0 to find the following:

a. the position function

b. velocity function

c. time required for object to hit ground

d. velocity of object at impact


Thank you!

Guest Nov 20, 2018

a    you are given the position function

b     velocity function is the derivative of position fxn

c         time will be when s(t) = 0

d       When you have THAT time , substitute that value of 't' in to the velocity function from 'b'

ElectricPavlov  Nov 20, 2018

s ' (t)   =  -32t + Vo



Vo  = 0     [ the object is dropped...so the initial velocity = 0 ]

So  =   initial height  = 64 ft


So we have


0   = -16t^2 + 64       add 16t^2 to both sides


16t^2   = 64       divide both sides by 16


t^2   = 4       take the positive root


t  = 2   (seconds to hit the ground)






Velocity at impact


-32(2)   =   -64 ft/ s



cool cool cool

CPhill  Nov 20, 2018

5 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.