your friend tosses a ball into the air at an initial velocity of 18 feet per second.The equation h=-8t^2+18t+5 models the height h of the ball t seconds after it was thrown.
When did the ball reach its highest point?
To find the time when the ball reaches its highest point, we need to determine the vertex of the quadratic equation h = -8t^2 + 18t + 5.
The equation is in the form h = at^2 + bt + c, where a = -8, b = 18, and c = 5.
The x-coordinate of the vertex can be found using the formula t = -b / (2a).
Plugging in the values, we have t = -18 / (2 * -8).
Simplifying further, we get t = -18 / -16 = 1.125.
Therefore, the ball reaches its highest point at 1.125 seconds after it was thrown.
To find the time when the ball reaches its highest point, we need to determine the vertex of the quadratic equation h = -8t^2 + 18t + 5. The vertex form of a quadratic equation is given by h = a(t - h)^2 + k, where (h, k) represents the coordinates of the vertex.
In this case, a = -8, so we need to find the values of h and k. We can use the formula:
h = -b / (2a)
k = c - (b^2 / (4a))
In the given equation, b = 18 and c = 5. Substituting these values, we get:
h = -18 / (2*(-8)) = -18 / -16 = 1.125
Now we substitute h back into the equation to find k:
k = 5 - (18^2 / (4*(-8))) = 5 - (324 / -32) = 5 + 10.125 = 15.125
Therefore, the vertex of the equation is (1.125, 15.125).
The x-coordinate of the vertex represents the time when the ball reaches its highest point. So, the ball reaches its highest point 1.125 seconds after it was thrown.