A ball is thrown vertically upward with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s(t)=96t-16t^2
a.) at what time will the ball strike the ground
b.) for what time t is the ball more than 128 feet above the ground?
c.) when will the ball reach its highest peak? how high is it above the ground?
a)
when is s = 0 ?
0 = t(96 - 16 t)
t = 0 of course, it starts at 0
t = 96/16 = 6 seconds
b)
when is s = 128 (two times, on the way up and on the way down)
128 = 96 t - 16 t^2
8 = 6 t - t^2
t^2 - 6 t + 8 = 0
(t-4)(t-2) = 0
so between 2 seconds and 4 seconds
c)where is the vertex of this parabola?
Well it is halfway in time between 2 seconds and four seconds, which is 3 seconds
it is also halfway in time between 0 seconds and 6 seconds, which of course is also at 3 seconds
so
s(3) = 96(3) - 16(9)
= 144 feet
a.) Let's solve for the time when the ball strikes the ground by setting s(t) equal to 0: 0 = 96t - 16t^2. This is a quadratic equation, so we can solve by factoring or using the quadratic formula. Solving it using the quadratic formula, we get t = (96 ± √(96^2 - 4(-16)(0)))/(2(-16)). This simplifies to t = (96 ± √(9216))/(32), which further simplifies to t = (96 ± 96)/(32). Therefore, t = (96 + 96)/(32) = 192/32 = 6 seconds. So the ball will strike the ground after 6 seconds.
b.) To find the time when the ball is more than 128 feet above the ground, we set s(t) greater than 128 and solve for t: 128 < 96t - 16t^2. Rearranging this inequality, we get -16t^2 + 96t - 128 > 0. We can simplify this by dividing the entire inequality by 16, giving us -t^2 + 6t - 8 > 0. Now, we can factor this quadratic expression: -(t - 4)(t - 2) > 0. From this, we can see that the expression will be greater than zero when 2 < t < 4. Therefore, the ball will be more than 128 feet above the ground between 2 and 4 seconds.
c.) To find when the ball reaches its highest peak, we can look for the vertex of the quadratic equation s(t) = 96t - 16t^2. The highest point is the maximum value of the quadratic equation. The formula for finding the x-coordinate of the vertex of a quadratic equation in the form ax^2 + bx + c is x = -b/2a. In this case, a = -16 and b = 96, so the time at which the ball reaches its highest peak is t = -96/(2*(-16)) = 96/32 = 3 seconds.
To find how high the ball is above the ground at its highest peak, we substitute t = 3 into the equation s(t) = 96t - 16t^2: s(3) = 96(3) - 16(3)^2 = 288 - 16(9) = 288 - 144 = 144 feet.
So, the ball reaches its highest peak at 3 seconds and is 144 feet above the ground at that time.
a) To find the time at which the ball strikes the ground, we need to determine when the distance s(t) is equal to 0.
s(t) = 96t - 16t^2
Setting s(t) to 0:
0 = 96t - 16t^2
We can factor out a common term of 16t:
0 = 16t(6 - t)
Now, we have two possibilities:
1) 16t = 0
2) 6 - t = 0
For the first possibility, t = 0.
For the second possibility, solving for t:
6 - t = 0
t = 6
Therefore, the ball strikes the ground at t = 0 and t = 6 seconds.
b) To find the time when the ball is more than 128 feet above the ground, we need to determine when s(t) is greater than 128.
s(t) = 96t - 16t^2
Setting s(t) greater than 128:
96t - 16t^2 > 128
Rearranging the equation and setting it equal to zero:
16t^2 - 96t + 128 > 0
Factoring out a common term of 16:
16(t^2 - 6t + 8) > 0
Now we have two possibilities for the inequality to hold true:
1) t^2 - 6t + 8 > 0
2) t^2 - 6t + 8 < 0
Considering the first possibility, solving for t:
t^2 - 6t + 8 > 0
(t - 2)(t - 4) > 0
This inequality is true when either both factors are positive or both factors are negative.
(t - 2) > 0 and (t - 4) > 0
t > 2 and t > 4
Taking the intersection of these solutions, t > 4.
Therefore, the ball is more than 128 feet above the ground for t > 4 seconds.
c) To find the time when the ball reaches its highest peak, we need to find the vertex of the parabolic equation s(t) = 96t - 16t^2.
The vertex of a parabola in the form of y = ax^2 + bx + c is given by the coordinates (-b/2a, f(-b/2a)).
For our equation, a = -16, b = 96, and c = 0.
The x-coordinate of the vertex is given by -b/2a:
x = -96/(2 * -16)
x = -96/-32
x = 3
To find the y-coordinate, we substitute t = 3 into the equation s(t):
s(3) = 96(3) - 16(3^2)
s(3) = 288 - 144
s(3) = 144
Therefore, the ball reaches its highest peak at t = 3 seconds and it is 144 feet above the ground.
To find the answers to these questions, we need to analyze the given equation for the height of the ball. Let's break down each part step by step.
The equation given is:
s(t) = 96t - 16t^2
a.) To find the time when the ball strikes the ground, we need to determine when the height (s) is equal to 0. When the ball hits the ground, its height will be 0. So we set s(t) = 0 and solve for t.
0 = 96t - 16t^2
Rearrange the equation:
16t^2 - 96t = 0
Factor out the common term:
16t(t - 6) = 0
Now we have two possibilities:
1) 16t = 0, which implies t = 0.
2) t - 6 = 0, which implies t = 6.
Therefore, the ball will strike the ground at t = 6 seconds.
b.) To find the time when the ball is more than 128 feet above the ground, we need to set s(t) greater than 128 and solve for t.
96t - 16t^2 > 128
Rearrange the equation:
16t^2 - 96t + 128 < 0
Factor out the common term:
16(t^2 - 6t + 8) < 0
Now we have a quadratic equation in factored form:
16(t - 4)(t - 2) < 0
This inequality represents the times when the ball is more than 128 feet above the ground. We need to find the values of t that satisfy this inequality.
Using a number line or table of values, we can see that the values of t that satisfy the inequality are t < 2 and t > 4.
Therefore, the ball is more than 128 feet above the ground when t < 2 seconds or t > 4 seconds.
c.) To find the time when the ball reaches its highest peak, we need to determine the vertex of the parabolic equation. The vertex represents the maximum height.
The equation for the height of the ball is in the form s(t) = -16t^2 + 96t.
Since the coefficient of t^2 is negative, the parabola opens downwards. The vertex is given by t = -b/2a, where a = -16 and b = 96.
t = -96/(2(-16)) = -96/-32 = 3
Therefore, the ball reaches its highest peak at t = 3 seconds.
To find how high it is above the ground at the peak, substitute t = 3 into the equation:
s(3) = -16(3)^2 + 96(3)
s(3) = -16(9) + 288
s(3) = -144 + 288
s(3) = 144
Therefore, the ball is 144 feet above the ground at its highest peak.