MATH SOLVE

2 months ago

Q:
# A pelican flying in the air over water drops a fish from the height of 30 feet. The distance of the fish from the water can be measured by the function f(t)=-16t^2+30, where is the time in t seconds. A seagull is flying below the pelican when the fish falls. If the seagull flies along a path represented by s(t)=-8t+15, will the seagull be able to catch the fish before the fish hits the water? Justify your answer. If so, find the time it is caught and the distance above the water at that time.

Accepted Solution

A:

Answer with Step-by-step explanation:We are given that a pelican flying in the air over water drops a fish from the height of 30 feet.Distance of fish from the water=f(t)=[tex]-16t^2+30[/tex]The seagull flies along a path=[tex]s(t)=-8t+15[/tex]We have to find the seagull will able to catch the fish before the fish hits the water.If so, then we have to find time when it is caught and distance of fish above water at that time. According to question when seagull catch the fish then both are at the same height from water[tex]f(t)=s(t)[/tex][tex]-16t^2+30=-8t+15[/tex][tex]16t^2-8t+15-30=0[/tex][tex]16t^2-8t-15=0[/tex][tex](4t-5)(4t+3)=0[/tex][tex]4t-5=0[/tex] and [tex]4t+3=0[/tex][tex]4t=5\implies t=\frac{5}{4}[/tex][tex]4t+3=0[/tex][tex]4t=-3[/tex][tex]t=-\frac{3}{4}[/tex]It is not possible because time cannot be negative.Therefore, [tex]t=\frac{5}{4}=1.25[/tex]sSubstitute the value of t in equation II.[tex]s(1.25)=-8(1.25)+15=5 ft[/tex]The seagull will met the fish when the fish at 5 ft in the air.So, the seagull catch the fish before hits the water.