Aball is thrown in the air from the top of a building. its height, in meters above ground, as a function of time, in seconds, is given by h(t)=−4.9x2+19.6x+24.
a. how long does does it take for the ball to reach its maximum height?
b. what is the maximum height of the ball?

a.2s b.43.6m

Step-by-step explanation:

h(t)=−4.9t2+19.6t+24

24m = inittial possition

-4.9 t2= -1/2 g t2

19.6t=  Vo t

as the initial velocity is positive, the ball is thrown up.

the maximum height of the ball is reached when the velocity is 0

V= Vo-gt=19.6m/s-9.8m/s2 t=0

t=19.6/9.8=2s

it takes 2seconds for the ball to reach its maximum height

h(t)=−4.9t2+19.6t+24

h(2)=-19.6m+39.2m+24=43.6m

the maximum height of the ball is 43.6m

what are you trying to solve here

step-by-step explanation:

x = 3 ± √(5+y)