When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -5t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in ft/sec and “k” is the initial height in feet (as if you were on top of a tower or building).
To find the maximum height the ball reaches, you can analyze the given equation S = -5t^2 + v*t + k.
The equation represents the height of the ball at any time "t" in seconds.
"S" represents the height in feet, "v" represents the initial velocity or speed of the ball in ft/sec, and "k" represents the initial height of the ball in feet.
To determine the maximum height, we need to find the vertex of the parabolic shape described by this equation.
The vertex of a parabola in the form of y = ax^2 + bx + c can be found using the formula:
x = -b / (2a)
y = f(x) = -(b^2 - 4ac) / (4a)
In our given equation, S = -5t^2 + vt + k, the coefficient of t^2 is -5, and the coefficient of t is v.
Using the formula, we get:
t_vertex = -v / (2*(-5))
t_vertex = -v / -10
t_vertex = v / 10
Now, substituting the value of t_vertex back into the equation S = -5t^2 + vt + k, we can find the maximum height (S_max) the ball reaches.
S_max = -5*(v/10)^2 + v*(v/10) + k
S_max = -5*(v^2/100) + v^2/10 + k
S_max = (v^2/10) - (v^2/100) + k
S_max = (10v^2 - v^2)/100 + k
S_max = 9v^2/100 + k
So, the maximum height the ball reaches is given by the equation S_max = 9v^2/100 + k, where "v" is the initial velocity and "k" is the initial height of the ball.