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!
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
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
Thanks alot guys! Both of your answers were very helpful.
Yes, there is a relatively simple and straightforward way to determine the maximum height reached by the ball.
To find the maximum height, we need to find the vertex of the quadratic equation h = -5t^2 + 10t + 15. The vertex represents the peak or maximum point of the parabola.
To find the vertex, we can use the formula t = -b/2a, where a, b, and c are the coefficients of the quadratic equation (in this case, a = -5, b = 10, and c = 15).
Substituting the values, we have t = -10/(2*-5) = -10/-10 = 1.
So, the ball reaches its maximum height at t = 1 second.
Next, we substitute this value of t back into the equation to find the corresponding maximum height.
h = -5(1)^2 + 10(1) + 15 = -5 + 10 + 15 = 20.
Therefore, the maximum height reached by the ball is 20 meters.
In summary:
1. Use the formula t = -b/2a to find the time at which the ball reaches the maximum height.
2. Substitute this value of t back into the equation to find the maximum height.
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