A rocket is fired into the air, and its height in meters at any given time t can be calculated using the formula h(t) = 1600 + 4.9t^2. Find the maximum height of the rocket and the time at which it occurs.

First, find the first derivative of the height function h(t):

h'(t) = d(1600 + 4.9t^2)/dt = 9.8t

To find the maximum height, set h'(t) = 0 and solve for t:

9.8t = 0
t = 0

The maximum height occurs at t = 0. Now, find the height at this time:

h(0) = 1600 + 4.9(0)^2 = 1600

The maximum height of the rocket is 1600 meters, and it occurs at time t=0.

To find the maximum height of the rocket and the time at which it occurs, we can use the given formula h(t) = 1600 + 4.9t^2. The maximum height is reached when the derivative of the function is equal to zero.

First, let's find the derivative of h(t) with respect to t:
h'(t) = 0 + 2 * 4.9 * t^(2-1)
= 9.8t

Next, let's set h'(t) equal to zero and solve for t:
9.8t = 0

Dividing both sides by 9.8, we get:
t = 0

Therefore, the time at which the rocket reaches its maximum height is t = 0.

To find the maximum height, we substitute this value of t into the original formula h(t):
h(0) = 1600 + 4.9(0)^2
= 1600 + 0
= 1600

Therefore, the maximum height of the rocket is 1600 meters.

To find the maximum height of the rocket and the time at which it occurs, we need to use calculus. We'll differentiate the equation to find the critical points and then analyze them to determine the maximum height.

Let's start by differentiating the equation h(t) = 1600 + 4.9t^2 with respect to time, t:

h'(t) = 0 + 2 * 4.9t

Simplifying this gives us:

h'(t) = 9.8t

Now, set h'(t) equal to zero and solve for t:

9.8t = 0

t = 0

This gives us the critical point at t = 0. However, since we are interested in the time when the rocket reaches its maximum height, we don't consider t = 0 because this corresponds to the initial time when the rocket was fired.

Next, we need to find the second derivative, which helps us determine whether the critical point we have found is a maximum or minimum:

h''(t) = 9.8

Since the second derivative, h''(t), is a constant (9.8), this indicates that the critical point at t = 0 is a minimum.

Now, let's consider the endpoints of the rocket's flight. The rocket is fired into the air, so time t starts from 0. However, we don't know how long the rocket will stay in the air. Let's call that time T until the rocket reaches the ground. Therefore, we need to consider the height at time T.

Plug in t = T into the original equation h(t) = 1600 + 4.9t^2:

h(T) = 1600 + 4.9T^2

To find the maximum height, we need to find the peak between t = 0 and t = T, so we compare the heights at time t = 0, t = T, and the critical point.

h(0) = 1600 + 4.9(0)^2 = 1600
h(T) = 1600 + 4.9T^2

Comparing these values, we can see that the rocket reaches the maximum height at time T.

Therefore, the maximum height of the rocket is h(T) = 1600 + 4.9T^2, and the time at which it occurs is t = T. However, to find the exact value of T, we need more information about the rocket's flight duration or when it reaches the ground.