Step-by-step explanation:
I've posted solutions in the picture. Rather the ways to solve them. Check and find answers on your own.
Also, I've not solved the problem by differential calculus. You can, obviously, if you're interested. Use it for ease and for tougher equations.
The solution is:
a) time to reach s = 288 ft is 3 sec
b) Total time of the fly is 9 sec
a) The height s = - 16×t² + v₀×t
At a height of 288 ft with v₀ = 144 ft/sec
288 = -16t² + 144×t or 16×t² - 144×t + 288 = 0
Simplifying dividing by 16 t² - 9×t + 18 = 0
We got a second degree equation, solving for t
t₁,₂ = [ 9 ±√81 - 72 ]/2
t₁ = (9 + 3)/2 t₁ = 6 sec and t₂ = ( 9 - 3 ) /2 t₂ = 3 sec
we got 2 possible solution, we will examine s maximum
when s is maximum ds/dt = 0 then
s = - 16×t² + v₀×t tacking derivatives on both sides of the equation
ds/dt = - 32× t + v₀ ds/dt = 0 = - 32× t + 144
- 32×t + 144 = 0 t = 144/32 t = 4,5 sec
And s max = - 16 ×( 4,5)² + 144 ( 4,5)
s maximum is s = - 324 + 648 s max = 324 ft
Then if s maximum will occurs at 4,5 sec and it was 324 ft, it is obviuos that time to get 288 ft was with t₂ = 3 sec.
b) As the time to get s maximum is 4,5 sec, by symmetry total time is 9 sec
