Question

The function h(0) = -16t2 + vt + 10 gives the height of a platform diver above the water, in feet, t seconds after

Answer 1

Complete Question:

The height of a diver jumping from a diving platform is about $h(t)= 16t^2 +11t+10$, where h is the height of the diver in feet above the water and t is the time measured in seconds, when diving from a platform about 34.3 feet above the water with an upward initial velocity of 11 ft/sec..  After how many seconds is the diver's height above the water 15 feet?

Answer:

0.3125 seconds

Step-by-step explanation:

Given

$h(t)= 16t^2 +11t+10$

Required

Time when the diver is above 15ft

This implies that h(t) = 15

So, we have:

$15= 16t^2 +11t+10$

Collect Like Terms

$0= 16t^2 +11t+10 - 15$

$0= 16t^2 +11t-5$

$16t^2 +11t-5=0$

Expand:

$16t^2 + 16t - 5t - 5 = 0$

Factorize:

$16t(t + 1) - 5(t + 1) = 0$

$(16t - 5)(t + 1)=0$

Split

$16t - 5 = 0$ or $t + 1 = 0$

$16t = 5$ or $t + 1 = 0$

Solve for t

$t = 5/16$ or $t = -1$

$t = 0.3125$ or $t = -1$

But time can't be negative. So:

$t = 0.3125$

Hence, it will take the diver 0.3125 seconds to be at a height above 15 feet

Answer 2

It will take t = 0.3125 seconds is the diver's height above the water 15 feet.

Given ,

The height of a diver jumping from a diving platform is h

 h(t) = $-16t^{2} + vt + 10$

The diver's height above the water 15 feet

Velocity = v = 11 ft per second

We have to find, t seconds is the diver's height above the water 15 feet.

According to the question,

$h(t) = -16t^{2} +vt+10$

Where v = 11, h(t) = 15

15 = -16$t^{2}$ + 11t + 10  

15 - 10 = $-16t^{2} + 11t$

5 = $-16t^{2} +11t$

$16t^{2} -11t +5= 0$

$16t^{2} -16t - 5t + 5 = 0$

16t ( t - 1 ) - 5 ( t + 1 ) = 0

(16t - 5 ) ( t + 1 ) = 0

16t - 5 = 0

16t = 5

t = $\frac{5}{16}$

t = 0.3125

And t + 1 = 0

t = -1

Time can not be negative so we can't take t = -1 .

So, The required value of time is t = 0.3125

Hence, it will take t= 0.3125 seconds is the diver's height above the water 15 feet.

For the more information about Equation of motion click the link given below.

https://brainly.com/question/13514745