A player throws a ball up and toward a wall that is 17 feet high. The height h in feet of the ball after t seconds after it leaves the player's hand is modeled by h=-16t^2+25t+6. If the ball makes it to where the wall is, will it go over the wall or hit the wall?
Find the vertex as well.
To determine whether the ball will go over the wall or hit the wall, we need to find the maximum height of the ball's trajectory.
The equation given is: h = -16t^2 + 25t + 6
The vertex of a quadratic equation in the form of h = at^2 + bt + c can be found using the formula: t = -b/2a
In this case, a = -16 and b = 25. Plugging these values into the formula, we have:
t = -25 / (2 * -16)
t = -25 / -32
t = 0.78125
This means that the maximum height is reached after approximately 0.78125 seconds.
Substitute this value back into the equation to find the maximum height:
h = -16(0.78125)^2 + 25(0.78125) + 6
h = -9.765625 + 19.53125 + 6
h = 15.765625
Therefore, the maximum height of the ball is approximately 15.765625 feet.
Since the wall is 17 feet high, the ball will go over the wall.
To determine if the ball will go over the wall or hit the wall, we need to find the maximum height reached by the ball. The vertex of the quadratic equation represents the highest point of the ball's trajectory.
The given equation that models the height of the ball is h = -16t^2 + 25t + 6. To find the vertex of this quadratic equation, we can use the formula t = -b / (2a), where a, b, and c are the coefficients of the quadratic equation (in the form ax^2 + bx + c).
By comparing the given equation to the standard form ax^2 + bx + c, we can determine that a = -16 and b = 25. Plugging these values into the formula, we get:
t = -25 / (2 * -16)
t = -25 / -32
t ≈ 0.78125
To find the corresponding height at that time, substitute t = 0.78125 into the equation:
h = -16(0.78125)^2 + 25(0.78125) + 6
h ≈ -9.515625 + 19.53125 + 6
h ≈ 16.015625
The vertex of the quadratic equation is approximately (0.78125, 16.015625).
Since the maximum height (16.015625 feet) is less than the height of the wall (17 feet), the ball will not go over the wall. It will hit the wall.
think back to your algebra I.
Where is the vertex of a parabola?
remember x = -b/2a ?