Question

A seagull flying in the air over water drops a crab from a height of 45 feet.The distance the crab is from the water as it falls can be represented by the function h(t)=-16t^2+45,where t is time measured in seconds.To catch the crab as it falls,a seagull flies along a path represented by the function g(t)=-13t+23

1. If the seagull catches the crab, then what height does it catch the crab? After how many seconds does it take for the seagull to catch the crab?

2. Sketch and label a coordinate plane to model the situtation.

Answer 1

For the seagull to catch the crab, h(t)=g(t) so:

-16t^2+45=-13t+23 add 16t^2 to both sides

45=16t^2-13t+23 subtract 45 from both sides

16t^2-13t-22=0 using the quadratic equation:

t=(13±√1577)/32, since t>0

t=(13+√1577)/32 seconds

t≈1.65 seconds (to nearest hundredth of a second)

And the height that this occurs using either original equation is:

h((13+√1577)/32)≈1.59 ft (to nearest hundredth of a foot)