The equation ℎ = −162 + 32 + 9 gives the height of a ball, h, in feet above the ground, at t seconds after the ball is thrown upward. How many seconds after the ball is thrown will it reach its maximum height? What is the max height?

time = 1secs

Max height = 25metres

Step-by-step explanation:

Let the equation that models the height of the ball be expressed as;

h(t) = −16t^2 + 32t + 9

t is the time in seconds

The velocity of the ball at its maximum height is zero i.e dh/dt = 0

dh/dt = -32t + 32

0 = -32t + 32

32t = 32

t = 32/32

t = 1 seconds

The ball will reach its maximum height after 1 seconds.

Get the maximum height reached. Substitute t = 1 into the modeled function;

h(t) = −16t^2 + 32t + 9

h(1) = −16(1)^2 + 32(1) + 9

h(1) = −16 + 32 + 9  

h(1)= 16+9

h(1) = 25

Hence the maximum height reached is 25metres

i did it myself and i got 4.2 (b) hope its right

step-by-step explanation:

step-by-step explanation:

let

we know that

the y-intercept is the y-coordinate of the point when the value of x is equal to zero

so

the y-coordinate of the point a is the y-intercept

the answer part a) is

the value of y-intercept is

part b) find the slope m

we know that

the value of the slope is equal to

substitute the values

therefore

the answer part b) is

the value of the slope is equal to

the general answer is

the slope is

the value of y-intercept is