A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function h = −16t2 + 36t + 5. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height?
The maximum height of course is the y value of the vertex of this downwards opening parabola, in the form (x,y)
for yours the x of the vertex is -36/-32 = 9/8 = 1.125 seconds
the actual max height is -16(9/8)^2 + 36(9/8) + 5 = 101/4 ft or 25.25 ft
There are several ways to do this question, another would be
to complete the square of the function, or if you know Calculus,
set the derivative equal to zero, to find the t when the max happens.
If this is an Algebra 2 problem, use x=-b/2a to find the x-coordinate of the vertex and then plug it in the function to find the y-coordinate.
If this is a Calculus problem, set the derivative equal to 0 to find the time when the function hits its maximum.
To find the time at which the ball reaches its maximum height, we need to determine the vertex of the parabolic function. In this case, the height of the ball is given by the function h = -16t^2 + 36t + 5.
The vertex of a parabola is given by the formula t = -b / (2a), where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c = 0. In this case, a = -16 and b = 36.
Substituting the values into the formula, we have t = -36 / (2*(-16)) = -36 / (-32) = 1.125 seconds.
Rounded to the nearest hundredth, the ball reaches its maximum height at approximately 1.13 seconds.
To find the maximum height of the ball, we substitute this time value back into the equation h = -16t^2 + 36t + 5:
h = -16(1.13)^2 + 36(1.13) + 5
Simplifying, we get h ≈ 25.36 feet.
Therefore, the ball reaches its maximum height in approximately 1.13 seconds, and the maximum height is approximately 25.36 feet.