Question

Hello,
Is there easiest and simplest way to do this question?

"A baseball is thrown from the top of a building and falls to the ground below. The height of the baseball above the ground is approximated by the relation h = -5t2 + 10t + 15, where h is the height above the ground in metres and "t" is the elapsed time in seconds. Determine the maximum height that is reached by the ball."

Thank you very much!

Answer 1

that is a parabola (quadratic)

5 t^2 - 10 t - 15 = -h

find the vertex by completing the square

t^2 - 2 t - 7.5 = -(1/5)(h)

t^2 - 2 t = -h/5 + 7.5

t^2 - 2 t + 1 = -h/5 + 8.5

(t-1)^2 = -(1/5) (h - 42.5)

so max height at one second of 42.5 meters

Answer 2

sure, all you have to do is to find the vertex of this "downwards" parabola.

I will assume that you do not know Calculus , so
(easiest way):

the x of the vertex for y = ax^2 + bx + c
is -b/(2a)
so for your equation, the
t of the vertex is -10/-10 = 1
sub into the equation ....
h = -5(1^2) + 10(1) + 15
= -5 + 10 + 15 = 20
vertex is (1,20)

so the ball reaches a max of 20 m after 1 second

Answer 3

t^2 - 2 t - 3 = -(1/5)(h)

t^2 - 2 t = -h/5 + 3

t^2 - 2 t + 1 = -h/5 + 4

(t-1)^2 = -(1/5) (h - 20)

so max height at one second of 20 meters

Answer 4

Thanks alot guys! Both of your answers were very helpful.