if h(t) represents the height of an object above ground level at time t and h(t) is given by h(t)=-16t^2+13t+1.
find the height of the object at the time whent eh speed is zero
Ah, the fascinating world of physics! Let's find the height of the object when the speed is zero.
To find the time when the speed is zero, we need to find the derivative of h(t) with respect to time t and set it equal to zero. Let's get fancy with some calculus. *picks up a monocle and a top hat*
The derivative of h(t) is obtained by differentiating each term. The derivative of -16t^2 is -32t, the derivative of 13t is 13, and the derivative of 1 is 0. Combining these, the derivative of h(t) is -32t + 13.
Alright, now we set -32t + 13 equal to zero and solve for t. Let me grab my trusty calculator and my clown nose. *places nose on face and calculates*
-32t + 13 = 0
-32t = -13
t = -13/-32
t ≈ 0.40625
Ta-da! The time when the speed is zero is approximately 0.40625. Now let's substitute this value back into the original equation to find the height.
h(0.40625) = -16(0.40625)^2 + 13(0.40625) + 1
After another round of calculations and some clown wiggles...
h(0.40625) ≈ 3.265625
The height of the object when the speed is zero is approximately 3.265625 units above ground level.
Hope I could bring a smile to your face with this answer!
To find the height of the object at the time when the speed is zero, we need to determine when the derivative of the height function is zero.
1. Take the derivative of the height function h(t) with respect to time t:
h'(t) = -32t + 13
2. Set h'(t) equal to zero and solve for t:
-32t + 13 = 0
-32t = -13
t = (-13)/(-32)
t = 13/32
3. Substitute the value of t into the height function h(t) to find the height at that time:
h(13/32) = -16(13/32)^2 + 13(13/32) + 1
h(13/32) = -16(169/1024) + 169/32 + 1
h(13/32) = -2704/1024 + 5408/1024 + 1024/1024
h(13/32) = 3705/1024
Therefore, the height of the object at the time when the speed is zero is 3705/1024.
To find the height of the object at the time when the speed is zero, we need to find the value of time, denoted as t, where the derivative of the height function, h'(t), equals zero.
First, let's differentiate the height function h(t) with respect to time t to find h'(t):
h'(t) = -32t + 13
Next, we'll set h'(t) equal to zero and solve for t:
-32t + 13 = 0
-32t = -13
t = -13 / -32
t = 13/32
So, the time when the speed is zero is t = 13/32.
To find the height of the object at this time, we substitute this value of t back into the height function h(t):
h(t) = -16t^2 + 13t + 1
h(13/32) = -16(13/32)^2 + 13(13/32) + 1
Now, we can solve this equation to find the height of the object at the specified time.
The velocity is represented by the first derivative of h(t)
so h'(t) = -32t + 13
you want that to be zero
so
-32t + 13 = 0
t = 13/32 seconds
plug that into the original and you are done