An object, at the top of a very tall building, is released from rest and falls freely due to gravity. Neglect air resistance and calculate the distance covered by the object between times t1 = 3.89 s and t2 = 6.10 s after it is released.

Question

anonymous · Answer

Please help me!

cwrw238 · Answer

use one of the equations of motion for a body falling under gravity

you need to find distance s , acceleration due to grsvity = 32 ft s^-2

s = (1/2)*32 * t^2
that is
s = 16t^2

cwrw238 · Answer

now  calculate the  value of s  when t = 3.89  and then calculate s when t = 6.10 
and subtract the last distance from the first to get your answer

cwrw238 · Answer

that is  

16*(6.10)^2 - 16* (3.89)^2

anonymous · Answer

thank you!

cwrw238 · Answer

yw