A rocket is launched into the air verically is tracked by a radar station on the ground 3 km from the launch site. What is the veritcal speed of the rocket at the instant when the distance from the radar station is 5 km and this distance is increasing at a rate of 5 000km/hr ???
The height of the rocket at that time is sqrt (5^2 - 3^2) = 4 km
Let x = distance to the rocket and h = height.
x^2 = h^2 + 9
2x dx/dt = 2h dh/dt
10 * 5000 = 8 * dh/dt
dh/dt = vertical speed = 50,000/8 = 6250 km/hr
To determine the vertical speed of the rocket at the instant when the distance from the radar station is 5 km, we can use the given information about the rate at which the distance is increasing.
First, let's derive the relationship between the distance and height of the rocket. We can use the Pythagorean theorem to find the height of the rocket at that time:
x^2 = h^2 + 9
Here, x represents the distance to the rocket and h represents the height. In this case, x = 5 km and h = sqrt(5^2 - 3^2) = 4 km.
Now, let's differentiate the equation with respect to time t to find the rates of change:
2x * dx/dt = 2h * dh/dt
Since we want to find the vertical speed of the rocket (dh/dt), let's focus on that term. We know that dx/dt (the rate at which the distance is increasing) is 5000 km/hr.
So, we have:
2 * 5 * 5000 = 2 * 4 * dh/dt
Simplifying the equation:
10000 = 8 * dh/dt
Finally, solve for dh/dt, which represents the vertical speed of the rocket:
dh/dt = 10000 / 8 = 6250 km/hr
Therefore, the vertical speed of the rocket at the instant when the distance from the radar station is 5 km, and this distance is increasing at a rate of 5000 km/hr, is 6250 km/hr.