Question

A parachutist jumps from an airplane and immediately opens her parachute. Her altitude in metres, after t seconds, is modelled by the equation y = -11t + 500. A second parachutist jumps 5 secs later and free falls for a few seconds. His altitude during this time is modeled by the equation y = -4.9 (t-4)^2 + 500. When does he catch up to the first parachutist? (HINT: When do they meet in the air?)

Answer 1

The second parachutist catches up to the first after approximately 10.245 seconds.

Step-by-step explanation:

To find when the second parachutist catches up to the first, we need to find the time when their altitudes are equal. We can set the two equations equal to each other and solve for t:

-11t + 500 = -4.9(t-4)^2 + 500

Expand and simplify this equation:

-11t + 500 = -4.9(t^2 - 8t + 16) + 500

Simplify further:

-11t + 500 = -4.9t^2 + 39.2t - 78.4 + 500

Combine like terms:

-4.9t^2 + 50.2t = 0

Factor out t:

t(-4.9t + 50.2) = 0

Set each factor equal to zero and solve for t:

t = 0 or t = 10.245

Since time cannot be negative in this context, the second parachutist catches up to the first after approximately 10.245 seconds.