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!
+36916 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'
+128460 b
s ' (t) = -32t + Vo
c
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)
d
Velocity at impact
-32(2) = -64 ft/ s
