quadratic equation

Question

A gun fires a bullet almost straight up from the edge of a 120-foot cliff. If the bullet leaves the gun with a speed of 476 feet per second, its height at time t is given by h(t)=-16t^2+476 t+120, measured from the ground below the cliff.

When will the bullet land on the ground below the cliff? (Hint: What is its height when it lands? Remember that we are measuring from the ground below, not from the cliff.)
The bullet will land after seconds.

Answer 1

h(t) = -16t² + 476t + 120

 

The height of the ground is h(t) = 0 feet.  So set h(t) = 0 and solve for t:

 

h(t) = -16t² + 476t + 120

0 = -16t² + 476t + 120

 

Divide both sides by 4:

 

0 = -4t² + 119t + 30

 

Use the quadratic formula to find the solutions:

 

t = 119/8 ± (1/8)√(119² - 4(-4)(30))

t = 119/8 ± 121/8

 

t = -1/4 and 30 seconds

 

The -1/4 seconds is the time it would have taken the bullet to reach a height of 120 feet with a velocity of 476 f/s at t = 0 had you fired it from the ground initially instead of from the 120 foot cliff.  The 30 seconds is the time it takes the bullet to reach the ground starting from 120 feet with an upward velocity of 476 f/s (being fired from the cliff).