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?

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.