If a ball is thrown vertically upward from the roof of 64 foot building with a velocity of 32 ft/sec, its height after t seconds is s(t)=64+32t–16t2. What is the maximum height the ball reaches "in ft"?
What is the velocity of the ball when it hits the ground (height 0) "in ft/sec"?
as you recall from algebra I, the vertex of a parabola occurs when t = -b/2a. In this case, that is when t = 1
Or, using calculus, since
ds/dt = 32-32t, ds/dt=0 when t=1
s=0 when t=1+√5, so plug that into ds/dt
To find the maximum height the ball reaches, we need to determine the vertex of the quadratic equation s(t) = 64 + 32t - 16t^2.
The vertex of a quadratic function in the form ax^2 + bx + c is given by the equation t = -b/2a.
In this case, a = -16 and b = 32.
Using the formula, we can find the time at which the ball reaches the maximum height:
t = -32 / (2 * -16)
t = -32 / -32
t = 1
So the ball reaches its maximum height 1 second after it was thrown.
To find the maximum height, we substitute t = 1 into the equation:
s(1) = 64 + 32(1) - 16(1^2)
s(1) = 64 + 32 - 16
s(1) = 80
Therefore, the maximum height the ball reaches is 80 ft.
To determine the velocity of the ball when it hits the ground (height 0), we can find the time at which the height is 0 by setting s(t) = 0:
0 = 64 + 32t - 16t^2
Rearranging the equation, we get:
16t^2 - 32t - 64 = 0
We can solve this quadratic equation using factoring or the quadratic formula. Factoring gives:
16(t^2 - 2t - 4) = 0
Setting each factor to zero, we can find the possible values of t:
t^2 - 2t - 4 = 0
Using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
Plugging in a = 1, b = -2, and c = -4, we can solve for t:
t = (-(-2) ± sqrt((-2)^2 - 4(1)(-4))) / (2(1))
t = (2 ± sqrt(4 + 16)) / 2
t = (2 ± sqrt(20)) / 2
t = (2 ± 2*sqrt(5)) / 2
t = 1 ± sqrt(5)
However, since the ball is thrown vertically upward, we are only interested in the positive value for t:
t = 1 + sqrt(5)
Therefore, the velocity of the ball when it hits the ground (height 0) is given by s(t) = 64 + 32t - 16t^2 with t = 1 + sqrt(5):
s(1 + sqrt(5)) = 64 + 32(1 + sqrt(5)) - 16(1 + sqrt(5))^2
Evaluating this expression will give us the velocity of the ball when it hits the ground (height 0) in ft/sec.