Question 198727
Simply set the height equations equal to each other and solve for "t"



{{{-16t^2 + 30=-8t + 15}}} Set the two right sides equal to each other.



{{{-16t^2+30+8t-15=0}}} Get all terms to the left side.



{{{-16t^2+8t+15=0}}} Combine like terms.



Notice we have a quadratic in the form of {{{At^2+Bt+C}}} where {{{A=-16}}}, {{{B=8}}}, and {{{C=15}}}



Let's use the quadratic formula to solve for "t":



{{{t = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{t = (-(8) +- sqrt( (8)^2-4(-16)(15) ))/(2(-16))}}} Plug in  {{{A=-16}}}, {{{B=8}}}, and {{{C=15}}}



{{{t = (-8 +- sqrt( 64-4(-16)(15) ))/(2(-16))}}} Square {{{8}}} to get {{{64}}}. 



{{{t = (-8 +- sqrt( 64--960 ))/(2(-16))}}} Multiply {{{4(-16)(15)}}} to get {{{-960}}}



{{{t = (-8 +- sqrt( 64+960 ))/(2(-16))}}} Rewrite {{{sqrt(64--960)}}} as {{{sqrt(64+960)}}}



{{{t = (-8 +- sqrt( 1024 ))/(2(-16))}}} Add {{{64}}} to {{{960}}} to get {{{1024}}}



{{{t = (-8 +- sqrt( 1024 ))/(-32)}}} Multiply {{{2}}} and {{{-16}}} to get {{{-32}}}. 



{{{t = (-8 +- 32)/(-32)}}} Take the square root of {{{1024}}} to get {{{32}}}. 



{{{t = (-8 + 32)/(-32)}}} or {{{t = (-8 - 32)/(-32)}}} Break up the expression. 



{{{t = (24)/(-32)}}} or {{{t =  (-40)/(-32)}}} Combine like terms. 



{{{t = -3/4}}} or {{{t = 5/4}}} Simplify. 



So <i>possible</i> the solutions are {{{t = -3/4}}} or {{{t = 5/4}}} 

  

Since a negative time doesn't make sense, this means that the only solution is {{{t = 5/4}}} which is {{{t=1.25}}}



So the pelican will catch the crab in 1.25 seconds. 



Notice how if we plug in {{{t=1.25}}} into either equation, we get: 


{{{g(1.25)=-8(1.25)+15=-10+15=5}}} which means that the pelican will meet with the crab 5 feet in the air. 



So the gull will catch the crab before the crab hits the water.