179k views
2 votes
A ball 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?

User RiccardoC
by
7.5k points

1 Answer

2 votes

Answer:

a.2s b.43.6m

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

User Jjreina
by
8.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.