SOLUTION: An object is thrown down from the top of a building 1280 feet tall with an initial velocity of 32 feet per second. The distance s (in feet) of the object from the ground after t se

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Question 1207823: An object is thrown down from the top of a building 1280 feet tall with an initial velocity of 32 feet per second. The distance s (in feet) of the object from the ground after t seconds is s = 1280- 32t- 16t^2.
(a) When will the object strike ground?
(b) What is the height of the object after 4 seconds?
Found 2 solutions by greenestamps, mananth:
Answer by greenestamps(13195) About Me  (Show Source):
You can put this solution on YOUR website!


(a) When it strikes the ground, its height is 0:

1280-32t-16t%5E2=0
80-2t-t%5E2=0
t%5E2%2B2t-80=0
%28t%2B10%29%28t-8%29=0

Reject the negative solution

t = 8

ANSWER: 8 seconds after it is thrown

(b) Substitute t=4 in the given equation

I leave that to you....

ANSWER: 896 feet

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
An object is thrown down from the top of a building 1280 feet tall with an initial velocity of 32 feet per second. The distance s (in feet) of the object from the ground after t seconds is s = 1280- 32t- 16t^2.
(a) When will the object strike ground?
(b) What is the height of the object after 4 seconds?
The equation

s=1280- 32t- 16t^2.
When object touches ground s=0
put s=0 and simplfy
t^2+2t-80=0
factorise
(t+10)(t-8)=0
t=-10 or t=8
The object hits the ground after 8 seconds
After 4 seconds
s=1280- 32t- 16t^2.
substitute t=4 and calculate
s=1280- 32*(4)- 16*4^2).=896 feet
The height of the object after 4 seconds is 896 feet.