A ball is thrown into the air with an upward velocity of 28 ft/s. Its height (h) in feet after t seconds is given by the function h = –16t² + 28t + 7. How long does it take the ball to reach its maximum height? What is the ball’s maximum height?
How do I solve this?
If you know Calculus, this is easy
dh/dt = -32t + 28 = 0 for a max of h
32t = 28
t = 28/32 = 7/8 = .875
h = -16(49/64) + 28(7/8) + 7 = 77/4 = 19.25 ft
if you don't know Calculus, you must complete the square
h = -16(t^2 - 28/16t + ...) + 7
= -16(t^2 - (7/4)t + 49/64 - 49/64) + 7
= -16( (t-7/8)^2 - 49/64) + 7
= -16(t-7/8)^2 + 49/4 + 7
= -16(t-7/8)^2 + 77/4
the vertex of this quadratic is (7,8 , 77,4)
so there is a max of h of 77/4 or 19.25 when h = 7/8
the vertex of a parabola is at
x = -b/2a = -28/-32 = 7/8
h(7/8) = 19.25
Ah, the classic ball-in-the-air problem. Let's solve it in a "jokingly" simple way.
To find the maximum height of the ball, we need to look at the equation h = -16t² + 28t + 7. Now, this equation happens to be in the form of a quadratic equation, where the coefficient of t² term is negative. This means that the graph of the equation will be a downward-facing parabola.
For a downward-facing parabola, the maximum height occurs at the vertex of the parabola. The x-coordinate of the vertex can be found using the formula:
t = -b / 2a
In our equation, a = -16 and b = 28. Plugging these values into the formula:
t = -28 / (2 * -16) = 28 / 32 = 7 / 8 ≈ 0.875
So, it takes approximately 0.875 seconds for the ball to reach its maximum height.
To find the maximum height itself, we can simply substitute this value of t back into the equation:
h = -16 * (0.875)² + 28 * (0.875) + 7
Simplifying the equation, we get:
h ≈ 20.125
Therefore, the ball's maximum height is approximately 20.125 feet.
And there you have it! The punchline to this problem is that the ball reaches its maximum height in about 0.875 seconds and that maximum height is approximately 20.125 feet. Ha-ha!
To find the time it takes for the ball to reach its maximum height, you need to determine when the velocity reaches zero. At the maximum height, the velocity becomes zero before the ball starts falling down due to gravity.
To find the time it takes for the ball to reach its maximum height, you need to find the vertex of the quadratic equation representing the height function h(t) = -16t^2 + 28t + 7.
The vertex of a parabola represented by the equation y = ax^2 + bx + c is given by the formula:
x = -b / (2a)
In this case, the equation is h(t) = -16t^2 + 28t + 7, so a = -16 and b = 28. Substituting these values into the formula, you get:
t = -28 / (2 * -16)
Simplifying further:
t = -28 / -32
t = 7/8
Therefore, it takes the ball 7/8 seconds to reach its maximum height.
To find the maximum height, substitute the value of t into the height function:
h = -16(7/8)^2 + 28(7/8) + 7
Simplifying further:
h = -16(49/64) + 28(7/8) + 7
h = -49/4 + 49/2 + 7
h = -49/4 + 98/4 + 28/4
h = 77/4
Therefore, the ball reaches a maximum height of 77/4 feet.
To solve this problem, we need to find the time at which the ball reaches its maximum height, as well as the value of the maximum height itself.
First, let's recall that the graph of a quadratic function (like the one given for the height of the ball) is a parabola. For a quadratic function in the form h = at² + bt + c, the parabola opens downwards if a < 0 and opens upwards if a > 0. In this case, the coefficient of the squared term is -16, so the parabola opens downwards.
The maximum height of the ball corresponds to the vertex of the parabola. To find the time at which the ball reaches its maximum height, we need to determine the x-coordinate of the vertex, using the formula: t = -b / (2a).
In the given equation, a = -16 and b = 28. Plugging these values into the formula, we have: t = -28 / (2 * -16) = -28 / -32 = 0.875 seconds.
Now, let's find the maximum height by substituting the value of t back into the original equation: h = –16t² + 28t + 7. Plugging in t = 0.875, we get: h = –16(0.875)² + 28(0.875) + 7. Evaluating this expression, we find: h ≈ 11.5 feet.
Therefore, it takes the ball approximately 0.875 seconds to reach its maximum height, and the maximum height of the ball is approximately 11.5 feet.