Physics: a ball is thrown vertically upward with an initial velocity of 80 feet per second. the distance (in feet) of the ball from the ground after t seconds if s=80t-16tt^2
a) draw the illustration
b) for what time interval is the ball more than 96 feet above the ground?
c) what is the maximum height of the ball?
d) after how many seconds does the ball reach the maximum height?
another great help!
a) To draw the illustration, we can create a graph with the vertical distance (s) on the y-axis and time (t) on the x-axis. The equation given is s = 80t - 16t^2.
b) To determine when the ball is more than 96 feet above the ground, we can set the equation s > 96 and solve for t.
80t - 16t^2 > 96
Rearranging, we get -16t^2 + 80t - 96 > 0
Factoring out -16, we have -16(t^2 - 5t + 6) > 0
Factoring further, we get -16(t - 3)(t - 2) > 0
Since the coefficient is negative, the inequality holds true when t lies between the roots of the equation. The roots are t = 2 and t = 3.
Therefore, the ball is more than 96 feet above the ground for the time interval (2, 3).
c) To find the maximum height of the ball, we need to find the vertex of the quadratic equation s = -16t^2 + 80t.
The vertex can be found using the formula t = -b/2a. In this case, a=-16 and b=80.
t = -80 / (2*(-16))
t = -80 / (-32)
t = 2.5
The maximum height occurs at t = 2.5. To find the height, we plug this value back into the equation:
s = -16(2.5)^2 + 80(2.5)
s = -16(6.25) + 200
s = -100 + 200
s = 100
Therefore, the maximum height of the ball is 100 feet.
d) The ball reaches the maximum height after 2.5 seconds.
a) To draw the illustration, we need to create a graph representing the distance of the ball from the ground over time. The x-axis will represent time (t in seconds), and the y-axis will represent the distance from the ground (s in feet).
b) To find the time interval during which the ball is more than 96 feet above the ground, we need to set the equation equal to 96 and solve for t.
80t - 16t^2 = 96
Rearranging the equation, we get:
16t^2 - 80t + 96 = 0
We can now factor the quadratic equation:
16(t^2 - 5t + 6) = 0
Factoring further, we have:
16(t - 2)(t - 3) = 0
Setting each factor equal to zero, we find two potential solutions:
t - 2 = 0 => t = 2 seconds
t - 3 = 0 => t = 3 seconds
Therefore, the ball is more than 96 feet above the ground between 2 and 3 seconds.
c) To find the maximum height of the ball, we need to determine the vertex of the parabolic equation. The vertex of a quadratic equation in the form "ax^2 + bx + c" can be found using the formula:
x = -b / 2a
In our equation, a = -16, b = 80. Plugging these values into the formula:
t = -(80) / (2 * -16)
t = -80 / -32
t = 2.5 seconds
Substituting this value for t back into the equation, we can find the maximum height:
s = 80(2.5) - 16(2.5)^2
s = 200 - 16(6.25)
s = 200 - 100
s = 100 feet
Therefore, the maximum height of the ball is 100 feet.
d) To find the time it takes for the ball to reach the maximum height, we use the t-value obtained in part c. The ball reaches the maximum height after 2.5 seconds.
b)
80t - 16t^2 > 96
16t^2 - 80t + 96 < 0
t^2 - 5t + 6 < 0
(t-2)(t-3) < 0
so the ball is higher than 96 for
2 < t < 3, that is, between 2 and 3 seconds
c) s = -16(t^2 - 5t + 25/4 - 25/4)
= -16(t-5/2)^2 + 100
I will leave it up to you to interpret that result.