Question

A stunt person falls from a helicopter 609 feet above sea level for an action movie and deploys a parachute when at a height of 500 feet above sea level. The stunt persons height above sea level is defined by the expression 600-16t^2, where t is the time , in minutes, after the fall.

At what time does the stunt person deploy the parachute ?

Answer 1

To find the time at which the stunt person deploys the parachute, we need to set the expression for their height above sea level equal to 500 feet and solve for t:

600 - 16t^2 = 500

Subtracting 500 from both sides:

100 - 16t^2 = 0

Dividing both sides by 16:

- t^2 = -100/16

Simplifying:

t^2 = 25/4

Taking the square root of both sides:

t = ±√(25/4) = ±(5/2)

Since time cannot be negative, we can discard the negative solution, leaving:

t = 5/2

Therefore, the stunt person deploys the parachute at a time of 5/2 minutes, which is equivalent to 2.5 minutes or 2 minutes and 30 seconds.