If a ball is thrown directly upward with a velocity of 80ft/s, its height (in feet) after t seconds is given by y=80t-16t^2. What is the maximum height attained by the ball?
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If a ball is thrown directly upward with a velocity of 80ft/s, its height (in feet) after t seconds is given by y=80t-16t^2. What is the maximum height attained by the ball?
The time to reach the maximum height and zero velocity derives from Vf = Vo - gt wherer Vf = the final velocity = 0, Vo = the initial velocity = 80fps, t = the time to zero velocity and g = the deleration due to gravity.
Therefore, 0 = 80 - 32t making t = 2.5 sec.
Then, from h = 80t-16t^2, h = 80(2.5) - 16(2.5)^2 = 200 - 100 = 100 ft.
That would be when the velocity is zero. That happens when the derivative
80 - 32t, is zero. Therefore t = 2.5 seconds at that time, and the maximum height is
80*2.5 - 16(2.5)^2 = 100 ft
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Well, let's find out together! To determine the maximum height attained by the ball, we need to determine the vertex of the parabolic equation y = 80t - 16t^2.
For a quadratic equation in the form y = ax^2 + bx + c, the x-coordinate of the vertex can be found using the formula x = -b / (2a).
In our case, a = -16 and b = 80. Plugging these values into the formula gives us x = -80 / (2 * -16) = 5.
Now that we have the x-coordinate of the vertex, we can substitute it back into the equation to find the y-coordinate. Plugging in t = 5, we get y = 80 * 5 - 16 * 5^2 = 400 - 400 = 0.
Therefore, the maximum height attained by the ball is 0 feet. However, don't worry, the ball will fall back down eventually!
To find the maximum height attained by the ball, we need to determine the vertex of the parabolic equation y = 80t - 16t^2.
The vertex form of a parabola is given by y = a(t-h)^2 + k, where (h,k) represents the coordinates of the vertex.
In this case, the equation y = 80t - 16t^2 represents a downward-opening parabola because of the negative coefficient of the t^2 term.
To find the vertex, we need to calculate the values of t, which is the time at which the ball reaches the maximum height.
Since the parabola is symmetric around its vertex, the maximum height is reached at the midpoint between the two roots of the equation when the ball hits the ground.
In this case, the ball hits the ground when y = 0. So we need to solve the equation 80t - 16t^2 = 0 to find the roots.
Factoring out t, we get t(80 - 16t) = 0.
Setting each factor to zero, t = 0 or 80 - 16t = 0.
Solving the second equation, we have 80 - 16t = 0.
Dividing both sides by 16, we get t = 5.
So the two roots of the equation are t = 0 and t = 5.
To find the time at which the ball reaches the maximum height, we take the average of the two roots: (0 + 5) / 2 = 2.5.
Substituting t = 2.5 into the equation y = 80t - 16t^2, we can find the maximum height:
y = 80(2.5) - 16(2.5)^2
y = 200 - 16(6.25)
y = 200 - 100
y = 100
Therefore, the maximum height attained by the ball is 100 feet.