If a ball is thrown vertically upward from the roof of 64 foot building with a velocity of 64 ft/sec, its height after t seconds is s(t)=64+64t–16t2 . What is the maximum height the ball reaches? What is the velocity of the ball when it hits the ground (height 0 )?
To find the maximum height the ball reaches, we need to find the vertex of the parabolic equation s(t) = 64 + 64t - 16t^2.
The vertex of a parabola in the form y = ax^2 + bx + c is given by the formula:
x = -b / (2a)
where in our case, a = -16 and b = 64.
Substituting these values into the formula, we get:
x = -64 / (2(-16))
x = -64 / (-32)
x = 2
Now let's find the corresponding height by plugging t = 2 back into the equation:
s(2) = 64 + 64(2) - 16(2)^2
s(2) = 64 + 128 - 16(4)
s(2) = 64 + 128 - 64
s(2) = 128
Therefore, the maximum height the ball reaches is 128 feet.
To find the velocity of the ball when it hits the ground (height 0), we need to find the time it takes for the ball to reach the ground.
Since the height of the ball at any time t is given by s(t), we can set s(t) equal to 0 and solve for t.
0 = 64 + 64t - 16t^2
Rearranging the equation:
16t^2 - 64t - 64 = 0
Dividing the equation by 16 to simplify it:
t^2 - 4t - 4 = 0
Now we can solve this quadratic equation for t using the quadratic formula:
t = (-(-4) ± √((-4)^2 - 4(1)(-4))) / (2(1))
t = (4 ± √(16 + 64)) / 2
t = (4 ± √80) / 2
t = (4 ± 4√5) / 2
t = 2 ± 2√5
Since time cannot be negative in this context, we take the positive value:
t = 2 + 2√5
Therefore, the time it takes for the ball to hit the ground is 2 + 2√5 seconds.
Now, to find the velocity of the ball when it hits the ground, we need to find the derivative of the height equation s(t) with respect to time t, and plug in the value of t.
s(t) = 64 + 64t - 16t^2
To find the derivative of s(t), we differentiate each term:
s'(t) = 0 + 64 - 16(2t)
s'(t) = 64 - 32t
Now we substitute t = 2 + 2√5 into the derivative equation to find the velocity at that time:
s'(2 + 2√5) = 64 - 32(2 + 2√5)
s'(2 + 2√5) = 64 - 64 - 64√5
s'(2 + 2√5) = -64√5
Therefore, the velocity of the ball when it hits the ground is -64√5 ft/sec.
To find the maximum height the ball reaches, we need to determine the vertex of the parabolic function s(t) = 64 + 64t - 16t^2.
The vertex of a parabola represented by the equation y = ax^2 + bx + c is given by the formula h = -b/2a, where h is the x-coordinate of the vertex.
In this case, the equation is s(t) = -16t^2 + 64t + 64, so we can identify a = -16 and b = 64.
Using the vertex formula, we can calculate the x-coordinate of the vertex:
t = -64 / (2 * -16)
= -64 / -32
= 2
Now we substitute this value of t back into the equation s(t) to find the height at that time:
s(2) = 64 + 64(2) - 16(2)^2
= 64 + 128 - 16(4)
= 64 + 128 - 64
= 128
Therefore, the maximum height reached by the ball is 128 feet.
To find the velocity of the ball when it hits the ground (height 0), we can use the derivative of s(t) with respect to t. The derivative will give us the rate of change of height with respect to time, which is the velocity.
Taking the derivative of s(t), we get:
s'(t) = 64 - 32t
To find the velocity when the ball hits the ground, we set s'(t) = 0:
64 - 32t = 0
Solving this equation for t gives us:
32t = 64
t = 2
So, the ball hits the ground after 2 seconds.
Now, we substitute this value of t back into s'(t) to find the velocity at that time:
s'(2) = 64 - 32(2)
= 64 - 64
= 0
Therefore, the velocity of the ball when it hits the ground is 0 ft/sec.
Differentiate to find the max height.
Velocity is 64 - 32t.
Max is at s'(t) = 0, when velocity is zero
s'(t)= 64 - 32t = 0
so it peaks at t=2. From that you can get the height.
You now need to find s(t) = 0, so solve 64+64t–16t2 = 0
That will give you the time, and from that the velocity function 64 - 32t will give you the answer to the second part