The height of an object is through downward with a velocity of 32 ft/seec from the roof of a 128 feet tall building is modeled by: h(t)=-16^2-32t+128

h=height of object off the ground (in feet)

t= seconds after the object is thrown.***How long does it take before the object his the ground?

Hello!

The object hits the ground when its height is zero.

h(t) = -16*t^2 - 32t + 128 = 0 -- it is a quadratic equation for t(note that the time t must be >=0).

Divide this by 16 to simplify:

-t^2 - 2t + 8 = 0, or t^2 + 2t - 8 = 0.

t_1,2 = (-2 +- sqrt(4 + 32))/2 = (-2 +- sqrt(36))/2 = (-1 +- 3) = -4 or 2.

Only 2(seconds) is suitable (>0). This is the answer.

OMG that was fast. I will totally pay you if you can help me with 3 more problems ?

This question appeared as "prepaid" and must be answered within 20 mins:) and it was simple.

You'll pay me for this problem after "3 more" or for 3 more separately?

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up