Answer :
Answer:
a) The maximum height is approx 15.5 unit.
b) The time it will take for Joey to reach the water is 1.45 hour.
Step-by-step explanation:
Given : When Joey dives off a diving board, the equation of his pathway can be modeled by [tex]h(t)= -16t^2+15t + 12[/tex]
To find : a) Find Joey's maximum height.
b) Find the time it will take for Joey to reach the water.
Solution :
Modeled [tex]h(t)= -16t^2+15t + 12[/tex] ....(1)
a) To find maximum height
Derivate (1) w.r.t. t,
[tex]h'= -32t+15[/tex]
For critical point put h'=0,
[tex]-32t+15=0[/tex]
[tex]t=\frac{15}{32}[/tex]
[tex]t=0.46875[/tex]
Derivate again w.r.t. t,
[tex]h''= -32<0[/tex]
It is maximum at t=0.46875.
Substitute t in equation (1),
[tex]h(0.46875)= -16(0.46875)^2+15(0.46875)+12[/tex]
[tex]h(0.46875)= -3.515625+7.03125+12[/tex]
[tex]h(0.46875)= 15.515625[/tex]
The maximum height is approx 15.5 unit.
b) To find the time it will take for Joey to reach the water.
Put h=0 in equation (1),
[tex]-16t^2+15t + 12=0[/tex]
Apply quadratic formula, [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Here, a=-16 , b=15, c=12
[tex]t=\frac{-15\pm\sqrt{(15)^2-4(-16)(12)}}{2(-16)}[/tex]
[tex]t=\frac{-15\pm\sqrt{225+768}}{-32}[/tex]
[tex]t=\frac{-15\pm\sqrt{993}}{-32}[/tex]
[tex]t=\frac{-15+\sqrt{993}}{-32},\frac{-15-\sqrt{993}}{-32}[/tex]
[tex]t=−0.515,1.453[/tex]
Reject negative value.
The time is t=1.45.
The time it will take for Joey to reach the water is 1.45 hour.