Question

Luis, using appropriate safety precautions, fires a .22 caliber gun directly into the sky above him. The bullet travels at a speed of 1200 feet per second. The following equation gives the height of the bullet at any point in time:
H (height) = -16t2 + 1200t, where t is the number of seconds; t = 0 is the instant when the gun is fired.
How long will the bullet be in the air?

Answer 1

The equation represents height as a function of time.

The bullet was in the air for 75 seconds.

The equation is given as:

$h(t) = -16t^2 + 1200t$

The time spent in the air is the time it takes the bullet to land on the ground.

When the bullet is on the ground, the height is:

$h(t) = 0$

So, we have:

$h(t) = -16t^2 + 1200t$

$-16t^2 + 1200t = 0$

Factor out t

$t(-16t + 1200) = 0$

Split

$t = 0$ or $-16t + 1200 = 0$

$t = 0$ represents when the shot is fired.

So, we solve for t in $-16t + 1200 = 0$

Collect like terms

$16t = 1200$

Divide both sides by 16

$t = 75$

Hence, the bullet was in the air for 75 seconds

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