the height of a ball thrown vertically upward from a rooftop is modelled by h(t)=-5t^2 +20t+50. Determine the maximum height of the ball.
To determine the maximum height of the ball, we need to find the highest point on the ball's trajectory, which corresponds to the vertex of the parabolic function.
The given height function is h(t) = -5t^2 + 20t + 50.
The vertex form of a parabolic function is h(t) = a(t - h)^2 + k, where (h, k) represents the vertex of the parabola.
By comparing the given height function to the vertex form, we can identify the values of a, h, and k.
In our case, a = -5 (the coefficient of t^2), h represents the time at the peak of the ball's trajectory, and k represents the maximum height reached by the ball.
To find the vertex, we can use the formula for the x-coordinate of the vertex: t = -b/(2a), where b represents the coefficient of t (in this case, 20).
Substituting the values into the formula, we have:
t = -20 / (2 * -5)
t = -20 / -10
t = 2
Therefore, the time at the peak of the ball's trajectory is 2 seconds.
Substituting this value back into the height function, we can find the maximum height:
h(t) = -5t^2 + 20t + 50
h(2) = -5(2)^2 + 20(2) + 50
h(2) = -5(4) + 40 + 50
h(2) = -20 + 40 + 50
h(2) = 70
Therefore, the maximum height of the ball is 70 units.
To determine the maximum height of the ball, we need to find the vertex of the parabolic function h(t) = -5t^2 + 20t + 50.
The vertex of a parabola in the form y = ax^2 + bx + c is given by the formula:
t = -b / (2a)
In this case, a = -5 and b = 20. Let's substitute these values into the formula:
t = -20 / (2*(-5))
t = -20 / (-10)
t = 2
Now, let's find the height at t = 2 by substituting it into the equation:
h(2) = -5*(2^2) + 20*2 + 50
h(2) = -5*4 + 40 + 50
h(2) = -20 + 40 + 50
h(2) = 70
Therefore, the maximum height of the ball is 70 units.