A ball is thrown up at the edge of a 364 foot cliff. The ball is thrown up with an initial velocity of 72 feet per second.
Its height measured in feet is given in terms of time t, measured in seconds by the equation h=−16t2+72t+364.
How high will the ball go?
To get how high the ball will go, am I supposed to factor −16t2+72t+364 or something?
Sorry, the equation is −16t^2+72t+364
nah. just recognize it's a parabola. Find the vertex. That will tell you how long it takes to get to its highest altitude. Then plug that value in for t in h(t) as given.
Finding the vertex of a parabola is easy, right. Algebra I.
oh ok. But if I wanted to find how long it takes the ball to come back to the ground then I would have to factor it right?
well, you could, or just use the Quadratic Formula. If you want to factor it, then we have
-16t^2+72t+364
-4(4t^2-18t-91)
Hmmm. Better stick with the QF.
thanks
To find how high the ball will go, we need to determine the maximum height reached by the ball.
The equation given for the height of the ball in terms of time is h = -16t^2 + 72t + 364, where h is the height in feet and t is the time in seconds.
To find the maximum height, we can use the fact that the maximum or minimum point of a quadratic function occurs at the vertex.
The vertex of a quadratic function represented by the equation y = ax^2 + bx + c can be found using the formula:
x = -b / (2a)
In our case, a = -16 and b = 72.
Substituting these values into the formula, we have:
t = -72 / (2 * -16)
t = -72 / (-32)
t = 2.25
So, the ball reaches its maximum height after 2.25 seconds.
Next, we substitute this value of t back into the equation to find the maximum height:
h = -16(2.25)^2 + 72(2.25) + 364
h = -16(5.06) + 162 + 364
h = -81 + 162 + 364
h = 445 feet
Therefore, the ball will reach a maximum height of 445 feet.