help me solve this please

Question

An object is dropped from 27 feet below the tip of the pinnacle atop a 703 ft tall building. Have a height h on the object of 30 seconds is given by the equation H=-16t²+672. Find how many seconds pass before the object reaches the ground

Answer 1

Ignoring typos and simply considering the equation...

Height (H) is 0 when the object reaches the ground

H = 0 = -16t2 + 672

Divide both sides by -16:
0 = t2 - 42
Move the constant to the others side:
42 = t2

Take the square root:
±6.48 = t
We can't have a negative time to land, so
t = 6.5, or 6 1/2 seconds to land

Answer 2

If the object is released from a height that is 27 ft below 703 ft, it must have been released from 703-27 = 676 ft, so the equation for its height should be

 

h(t) = -16t² + 676

 

Where did the 672 come from?

 

What do you mean by "Have a height h on the object of 30 seconds..." ?  The equation for h(t) applies from the moment the object is released until it hits the ground.

 

To find how many seconds after release the object hits the ground, set h(t) = 0 and solve for t.