Aball is thrown upward and its height after t seconds can be described by formula f(t)=−10t^2+24t+5.6. find the maximum height the ball will reach.
answer!

The maximum height reached by ball while throwing upward is 20 unit

Step-by-step explanation:

Given as :

The position of ball moving upward with time t sec is

f(t) = - 10 t² + 24 t + 5.6

Let the maximum height does the ball reach = H unit

Now, For maximum condition

= 0

So, = 0

Or, +   +   = 0

Or, ( - 10 ) × 2 t + 24 + 0 = 0

Or, - 20 t + 24 = 0

Or, 20 t = 24

∴   t =

i.e t =

Or, t = 1.2 sec

Again

Double differentiation of the function

i.e

Or, -20 + 0

i.e - 20

∵ - 20 0

So, The maximum height is gained at t = 1.2 seconds

So, Maximum height  at ( t = 1.2 sec) = - 10 t² + 24 t + 5.6

Or, H = - 10 (1.2)² + 24 × 1.2 + 5.6.

Or, H = - 1.44 × 10 + 28.8 + 5.6

Or, H = - 14.4 + 28.8 + 5.6

∴   H = - 14.4 + 34.4

i.e H = 20 unit

So, The maximum height = H = 20 unit

Hence, The maximum height reached by ball while throwing upward is 20 unit . Answer

it is a. 4.2 because 7.6-3.4 gets the remaining length of 4.2. i might be wrong but i am pretty sure this is the answer.

y=-2.5x+12.5

step-by-step explanation:

5y=2x+10

y=0.4x+2

slope=-2.5

5=-2.5*3+b

b=12.5