If a ball is thrown vertically upward from the roof of 48 foot building with a velocity of 96 ft/sec, its height after t seconds is s(t)=48+96t−16t^2. What is the maximum height the ball reaches?
What is the velocity of the ball when it hits the ground (height 0)?
the max height is on the axis of symmetry of the parabola
... t = -b / 2a ... -96 / (2 * -16)
... plug into the equation to find s
velocity is ds/dt
... solve for t when s is zero
... plug into the 1st derivative of s(t)
To find the maximum height the ball reaches, we need to find the vertex of the parabolic function s(t) = 48 + 96t - 16t^2.
The vertex of a parabola in the form of y = ax^2 + bx + c is given by the formula:
t = -b / (2a)
In this case, a = -16 and b = 96. Let's substitute these values into the formula to find the time at which the ball reaches its maximum height.
t = -(96) / (2 * -16)
t = -96 / -32
t = 3
So, the maximum height of the ball is reached at 3 seconds.
To find the velocity of the ball when it hits the ground (height 0), we need to find the value of t when s(t) = 0.
s(t) = 48 + 96t - 16t^2
Setting s(t) to zero:
0 = 48 + 96t - 16t^2
Rearranging the equation:
16t^2 - 96t - 48 = 0
We can solve this quadratic equation by factoring or by using the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 16, b = -96, and c = -48. Substituting these values into the quadratic formula:
t = (-(-96) ± √((-96)^2 - 4 * 16 * -48)) / (2 * 16)
t = (96 ± √(9216 + 3072)) / 32
t = (96 ± √(12288)) / 32
t = (96 ± 110.859) / 32
Using the positive root:
t = (96 + 110.859) / 32
t = 206.859 / 32
t ≈ 6.46
Therefore, the velocity of the ball when it hits the ground is approximately 6.46 seconds.
To find the maximum height the ball reaches, we need to determine the vertex of the parabolic equation s(t) = 48 + 96t - 16t^2. The vertex of a parabola in the form of ax^2 + bx + c is given by the formula: t = -b / (2a).
In this case, a = -16 and b = 96. Plugging these values into the formula, we get:
t = -96 / (2 * -16)
t = -96 / -32
t = 3
So, the ball reaches its maximum height after 3 seconds.
To find the maximum height, we substitute this value of t into the equation s(t):
s(3) = 48 + 96(3) - 16(3^2)
s(3) = 48 + 288 - 16(9)
s(3) = 48 + 288 - 144
s(3) = 192
Therefore, the maximum height the ball reaches is 192 feet.
Now, to find the velocity of the ball when it hits the ground (height 0), we can use the derivative of the position function s(t) with respect to time (t).
The derivative of s(t) is given by: v(t) = s'(t) = 96 - 32t.
To find the time when the ball hits the ground (height 0), we need to solve the equation s(t) = 0:
0 = 48 + 96t - 16t^2
Rearranging the equation, we get:
16t^2 - 96t - 48 = 0
Dividing both sides by 16, we get:
t^2 - 6t - 3 = 0
This equation can be solved using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values a = 1, b = -6, and c = -3, we can calculate the roots.
t = (-(-6) ± √((-6)^2 - 4(1)(-3))) / (2(1))
t = (6 ± √(36 + 12)) / 2
t = (6 ± √(48)) / 2
t = (6 ± √(16 * 3)) / 2
t = (6 ± 4√3) / 2
Simplifying further, we get:
t = 3 ± 2√3
Since we are looking for the time when the ball hits the ground, we are interested in the positive root:
t = 3 + 2√3
Now, we can substitute this value of t into the velocity function:
v(t) = 96 - 32 * (3 + 2√3)
v(t) = 96 - 96√3
Therefore, the velocity of the ball when it hits the ground is 96 - 96√3 ft/sec.
time in flight: solve s(t) = 0
max height at the vertex: t=3
v = 96-32t