A ball is thrown into the air with an upward velocity of 32 ft/s. Its height h in feet after t seconds is given by the function h = −16t2 + 32t + 0. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary.

Question

A ball is thrown into the air with an upward velocity of 32 ft/s. Its height h in feet after t seconds is given by the function h = −16t2 + 32t + 0. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary.

A: 1s; 10ft
B: 2s; 10ft
C: 1s; 26ft
D: 1s; 58ft

Answer 1

To find the time at which the ball reaches its maximum height, we need to determine the vertex of the function.

The given function is h = -16t^2 + 32t + 0, which represents the height of the ball at any given time t.

Since the coefficient of the t^2 term is negative, we know that the graph opens downwards, indicating that it represents a parabola.

The maximum height of the ball occurs at the vertex of the parabola, which is given by the formula: t = -b / (2a), where a and b are the coefficients of the quadratic equation.

In this case, a = -16 and b = 32. Plugging these values into the formula, we get:
t = -32 / (2 * -16) = -32 / (-32) = 1

Therefore, the ball reaches its maximum height at t = 1 second.

We can confirm this by substituting this value back into the equation:
h = -16(1)^2 + 32(1) + 0 = -16 + 32 + 0 = 16

Hence, the maximum height the ball reaches is 16 feet.

Therefore, the correct answer is option C: 1s; 26ft

Answer 2

To find the maximum height, we need to determine the vertex of the quadratic function h = −16t^2 + 32t + 0.

The vertex of a quadratic function in the form h = at^2 + bt + c is given by the formula t = -b / (2a).

For this problem, a = -16 and b = 32. Substituting these values into the formula yields:
t = -32 / (2 * -16)
t = -32 / -32
t = 1

So, the ball reaches its maximum height in 1 second.

To find the maximum height, substitute this value of t back into the original function:
h = −16(1)^2 + 32(1) + 0
h = -16 + 32 + 0
h = 16

Therefore, the ball reaches a maximum height of 16 feet.

The correct answer is:
D: 1s; 58ft

Answer 3

You will need the vertex.

Don't know which method you learned, but the simplest is this:
for y = ax^2 + bx + c,
the x of the vertex is -b/(2a)
= -32/-32 = 1

after 1 second, h = -16 + 32 = 16
So the max of 16 m is obtained after 1 second

or

compete the square:
h = -16(t^2 - 2t)
= -16(t^2 - 2t + 1 - 1)
= -16( (t-1)^2 - 1)
= -16(t-1)^2 + 16
so the vertex is (1,16), same result as above

unless you have a typo, none of the given choices are correct