Question

2. When Joey dives off a diving board, the equation of his pathway can be modeled by h = -16- + 15 + 12.

Answer 1

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 $h(t)= -16t^2+15t + 12$

To find : a) Find Joey's maximum height.

b) Find the time it will take for Joey to reach the water.

Solution :

Modeled  $h(t)= -16t^2+15t + 12$ ....(1)

a) To find maximum height

Derivate (1) w.r.t. t,

$h'= -32t+15$

For critical point put h'=0,

$-32t+15=0$

$t=\frac{15}{32}$

$t=0.46875$

Derivate again w.r.t. t,

$h''= -32$

It is maximum at t=0.46875.

Substitute t in equation (1),

$h(0.46875)= -16(0.46875)^2+15(0.46875)+12$

$h(0.46875)= -3.515625+7.03125+12$

$h(0.46875)= 15.515625$

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),

$-16t^2+15t + 12=0$

Apply quadratic formula, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here, a=-16 , b=15, c=12

$t=\frac{-15\pm\sqrt{(15)^2-4(-16)(12)}}{2(-16)}$

$t=\frac{-15\pm\sqrt{225+768}}{-32}$

$t=\frac{-15\pm\sqrt{993}}{-32}$

$t=\frac{-15+\sqrt{993}}{-32},\frac{-15-\sqrt{993}}{-32}$

$t=−0.515,1.453$

Reject negative value.

The time is t=1.45.

The time it will take for Joey to reach the water is 1.45 hour.