If h(t) represents the height of an object above ground level at time t and h(t) is given by h(t) = −16t2 + 96t + 1,
find the height of the object at the time when the velocity is zero.
take the first derivative and set it equal to zero.
To find the height of the object when the velocity is zero, we need to first find the time when the velocity is zero, and then substitute that time into the equation to find the corresponding height.
The velocity of an object can be obtained by taking the derivative of the height function with respect to time, so let's differentiate the given function h(t) = -16t^2 + 96t + 1 with respect to t:
h'(t) = -32t + 96
Now, we set the velocity function h'(t) equal to zero to find the time when the velocity is zero:
-32t + 96 = 0
Adding 32t to both sides gives:
32t = 96
Dividing both sides by 32:
t = 96/32
Simplifying further, we have:
t = 3
Therefore, the time when the velocity is zero is t = 3.
Now, we substitute this value of t back into the height function h(t) to find the corresponding height:
h(3) = -16(3)^2 + 96(3) + 1
Simplifying further:
h(3) = -16(9) + 288 + 1
h(3) = -144 + 288 + 1
h(3) = 145
Thus, the height of the object at the time when the velocity is zero is 145 units above the ground level.