A football player kicks the ball with an initial upward velocity of 60 feet per second. This situation is modeled using the equation h=−16t^2+60t+0
, where h is the height in feet and t is the time in seconds. Using this equation, what is the maximum range the ball will reach? Round to the nearest hundredth.(1 point)
I apologize for the confusion in my previous response. I made an error in the final calculation steps. Let's recalculate the maximum height correctly.
We already found the time at which the maximum height is reached (t = 1.875 seconds). Now let's plug that time back into the height equation to find the maximum height (h):
\[ h = -16t^2 + 60t \]
\[ h = -16(1.875)^2 + 60 \times 1.875 \]
\[ h = -16(3.515625) + 112.5 \]
\[ h = -56.25 + 112.5 \]
Here is the correction:
\[ h = -16 \times 3.515625 + 112.5 \]
\[ h = -56.25 \text{ (rounded from } -56.250000 \text{)} \]
\[ h = 112.5 - 56.25 \]
\[ h = 56.25 \text{ feet} \]
This is the correct maximum height that the ball will reach, which is 56.25 feet.
how many feet
To find the maximum range or the maximum height (h) the ball will reach, we can determine the time at which this maximum height occurs. Since the equation for height is a quadratic in the form of \( h=-16t^2+60t \), we can analyze it as a parabola opening downwards (because of the negative coefficient for the \( t^2 \)) term.
To find the time when the maximum height is reached, we can calculate the vertex of the parabola, which gives us the time value. For a quadratic equation in the form \( ax^2 + bx + c \), the t-coordinate of the vertex can be found with the formula \( t = -\frac{b}{2a} \).
In your equation, \( a = -16 \) and \( b = 60 \), so:
\[ t = -\frac{60}{2 \times (-16)} \]
\[ t = -\frac{60}{-32} \]
\[ t = \frac{60}{32} \]
\[ t = 1.875 \, \text{seconds} \]
This is the time at which the ball reaches its maximum height. To find the maximum height itself, insert this time into the equation:
\[ h = -16(1.875)^2 + 60 \times 1.875 \]
\[ h = -16 \times 3.515625 + 112.5 \]
\[ h = -56.25 + 112.5 \]
\[ h = 56.25 \, \text{feet} \]
This is the maximum height that the ball will reach, rounded to the nearest hundredth of a foot. Please note that in your original question, you asked for the maximum range, which typically refers to the horizontal distance traveled. However, the given equation only models height as a function of time, so we are only able to calculate the maximum height achieved, not the horizontal distance. If you are looking for how far the ball travels horizontally, we would need additional information about the ball's horizontal velocity or any horizontal forces acting on it.
To find the maximum range of the ball, we need to determine the horizontal distance traveled by the ball before it hits the ground. In this case, the equation h = -16t^2 + 60t + 0 describes the height of the ball as a function of time.
To find the range, we need to find the time at which the ball hits the ground. The ball hits the ground when its height, h, is equal to zero. Therefore, we can set the equation h = -16t^2 + 60t + 0 equal to zero and solve for t.
-16t^2 + 60t + 0 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, we can factor it as follows:
-16t(t - 3.75) = 0
Setting each factor equal to zero gives us:
-16t = 0 or t - 3.75 = 0
From the first equation, we get t = 0. This represents the initial time when the ball is kicked and has not yet started its upward journey. We can ignore this solution since we are interested in the time when the ball hits the ground.
From the second equation:
t - 3.75 = 0
t = 3.75
Therefore, the ball hits the ground at t = 3.75 seconds.
To find the range, we need to determine the horizontal distance traveled by the ball in this time. We can use the equation r = d/t, where r is the range, d is the distance traveled, and t is the time.
The distance traveled by the ball can be calculated using the equation d = 60t. Plugging in t = 3.75, we get:
d = 60(3.75)
d = 225
Therefore, the maximum range the ball will reach is 225 feet.