An object is projected directly upward from the ground. After t seconds it’s distance in feet above the ground is s=144t-16t^2
How long does it take the object to reach its maximum height?
What is the maximum height the object reaches? When does it hit the ground?
as with all quadratics ax^2+bx+c, the vertex (max or min) is at -b/2a
The max height is c - b^2/4a
since s = 16t(9-t) the roots (where s=0, that is, the object is on the ground) are at t=0, 9
don't forget your general knowledge about parabolas when you are confronted by the equation ...
Well, well, well... it seems like we have a little physics problem here, don't we? Don't worry, I'm here to make it fun!
To find the time it takes for the object to reach its maximum height, we need to find when the velocity of the object becomes zero. Why? Because at the maximum height, the object momentarily stops before changing direction and coming back down. It's like taking a little break before the ultimate descent!
So, let's differentiate that equation to find the velocity function. The derivative of s(t) with respect to t gives us v(t), which is the velocity function. Here's the good news: the formula for velocity is as simple as a popsicle stick. It's just the derivative of the distance function!
v(t) = 144 - 32t
Now, let's set v(t) equal to zero and solve for t:
144 - 32t = 0
Dividing both sides by 32 gives us t = 4.5 seconds. And that, my friend, is how long it takes for our little object to reach maximum height! Isn't it thrilling?
Now, let's solve for the maximum height. To do that, we need to plug t = 4.5 back into the distance function:
s(t) = 144t - 16t^2
s(4.5) = 144(4.5) - 16(4.5)^2
s(4.5) = 324 - 324
Well, well, well... it looks like the maximum height is a big, fat zero! Which means our object reaches a whopping height of zero feet at its peak. I guess it just wanted to take a brief vacation from being on the ground!
Now, let's find the time it takes for the object to hit the ground. We need to find when the distance function s(t) becomes zero:
144t - 16t^2 = 0
We can factor this quadratic equation as 16t(9 - t) = 0
So, either 16t = 0 (which we discard because it would mean the object never moved), or 9 - t = 0
Solving 9 - t = 0 gives us t = 9 seconds. And voilà, in 9 seconds, our little object will hit the ground like a graceful swan! I hope it sticks the landing!
To find the time it takes for the object to reach its maximum height, we need to find the time when the object's velocity becomes zero. The velocity can be calculated by taking the derivative of the equation for distance (s) with respect to time (t).
1. Differentiate the equation for s with respect to t:
ds/dt = 144 - 32t
2. Set ds/dt equal to zero and solve for t:
144 - 32t = 0
32t = 144
t = 144/32
t = 4.5 seconds
Therefore, it takes the object 4.5 seconds to reach its maximum height.
To find the maximum height, substitute the value of t into the original equation for s:
s = 144t - 16t^2
s = 144(4.5) - 16(4.5)^2
s = 648 - 16(20.25)
s = 648 - 324
s = 324 feet
So, the maximum height the object reaches is 324 feet.
To find when the object hits the ground, we need to find the time when the distance (s) is equal to zero.
1. Set the equation for s equal to zero:
144t - 16t^2 = 0
2. Factor out a common term:
16t(9 - t) = 0
3. Set each factor equal to zero and solve for t:
16t = 0 or 9 - t = 0
t = 0 or t = 9
Since time cannot be negative, t = 0 is not a valid solution.
Therefore, the object hits the ground after 9 seconds.
To find the time it takes for the object to reach its maximum height, we need to determine when the velocity of the object becomes zero. At maximum height, the velocity will be zero because the object momentarily stops before falling back down.
The velocity of the object is the derivative of its position with respect to time. Taking the derivative of the position equation s=144t-16t^2 gives us the velocity equation:
v = 144 - 32t
To find when the velocity becomes zero, we set v = 0 and solve for t:
144 - 32t = 0
32t = 144
t = 4.5 seconds
So, it takes 4.5 seconds for the object to reach its maximum height.
To find the maximum height of the object, we substitute the value of t into the position equation:
s = 144t - 16t^2
s = 144(4.5) - 16(4.5)^2
s = 648 - 324
s = 324 feet
Therefore, the maximum height the object reaches is 324 feet.
To find when the object hits the ground, we set the position equation equal to zero and solve for t:
144t - 16t^2 = 0
t(144 - 16t) = 0
t = 0 or t = 9
Since time cannot be negative, the object hits the ground after 9 seconds.
In summary:
- The object takes 4.5 seconds to reach its maximum height.
- The maximum height of the object is 324 feet.
- The object hits the ground after 9 seconds.