ball thrown vertically upward with an initial velocity of 80 ft per second. The distance s(in ft) of the ball from the ground after t seconds is s=80t-16t^2.
a. draw the illustration.
b. for what time interval is ball more than 96 ft above ground?
c. what's the max height of ball?
d. after how many seconds does the ball reach max height?
b. d = 80t - 16t^2 = 96
-16t^2 + 80t - 96 = 0
Divide both sides by 16:
-t^2 + 5t - 6 = 0
Use Quadratic Formula and get:
t = 2s, and 3 s.
Interval: 2< T <3.
c. h = (V^2-Vo^2)/2g
h = (0-80^2)/64 = 100 Ft.
d. V = Vo + g*t = 0 at max ht.
t = -Vo/g = -80/-32 = 2.50 s.
a. To create an illustration, we can plot the trajectory of the ball on a graph. The x-axis represents time in seconds (t), and the y-axis represents the distance from the ground in feet (s). The equation for the distance is given as s = 80t - 16t^2.
b. To find the time interval when the ball is more than 96 feet above the ground, we need to set up an inequality. We want to find the values of t that make s > 96. So, we have the inequality 80t - 16t^2 > 96. We can rearrange this equation to get -16t^2 + 80t - 96 > 0. To solve this inequality, we can factor it or use the quadratic formula.
c. The maximum height of the ball can be determined by finding the vertex of the parabolic function. The vertex is given by the formula t = -b / 2a, where a, b, and c are the coefficients of the quadratic equation. In this case, a = -16 and b = 80. Plugging in these values, we get t = -80 / (2 * -16) = 5 seconds. To find the maximum height, we can substitute this value of t back into the equation s = 80t - 16t^2.
d. From the previous step, we found that the ball reaches its maximum height after 5 seconds, as determined by the value of t at the vertex of the parabolic function.
a. To draw the illustration, you can create a graph with the vertical distance from the ground (s) on the y-axis and the time (t) on the x-axis. The equation given is s = 80t - 16t^2.
b. To find the time intervals when the ball is more than 96 ft above the ground, we can set the equation s = 96 and solve for t.
80t - 16t^2 = 96
Rearranging the equation, we get:
16t^2 - 80t + 96 = 0
Now we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, let's use factoring to solve it.
16t^2 - 80t + 96 = 0
Dividing the equation by 16 to simplify it, we get:
t^2 - 5t + 6 = 0
Factoring the equation, we get:
(t - 2)(t - 3) = 0
Setting each factor to zero, we find that t = 2 and t = 3.
Therefore, the ball is more than 96 ft above the ground during the time interval from t = 2 seconds to t = 3 seconds.
c. The maximum height of the ball can be determined by finding the vertex of the quadratic equation. The vertex is given by the formula t = -b / (2a), where a, b, and c are the coefficients of the quadratic equation.
In this case, a = -16, b = 80, and c = 0, so the equation becomes:
t = -80 / (2(-16))
Simplifying, t = -80 / (-32) = 2.5
To find the maximum height, substitute this value of t back into the equation s = 80t - 16t^2:
s = 80(2.5) - 16(2.5)^2 = 100 - 16(6.25) = 100 - 100 = 0
Therefore, the maximum height of the ball is 100 ft.
d. The time it takes for the ball to reach its maximum height can be found using the same formula as before: t = -b / (2a). In this case, a = -16 and b = 80.
t = -80 / (2(-16)) = 2.5 seconds
Therefore, the ball reaches its maximum height after 2.5 seconds.