Projectile motion radar antenna

[physics0926]

Homework Statement

A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the satellite to be 162 km away. The radar antenna is pointing upward at an angle of 62.3 degrees from the ground. Find the x and y component(in km) of the position of the satellite.

Homework Equations

Vx=dx/t ..... a=Vy-V0y/t ... Vy+V0y/2 = dy/t

The Attempt at a Solution

its a simple question but I am thinking too hard

V0y=Vo (cos) angle
Vx= Vo(cos) angle
dx?

 

[Kurdt]

Its just working out the lengths of two sides of a right angled triangle given the hypotenuse and angle. Draw a diagram of the situation and it will probably be a bit clearer.

 

[thomate1]

First, let's define our variables:
- V0y is the initial velocity in the y-direction (upward)
- Vx is the velocity in the x-direction (horizontal)
- angle is the angle at which the radar antenna is pointing
- dx is the horizontal distance from the antenna to the satellite
- dy is the vertical distance from the antenna to the satellite

Using the given information, we can solve for the vertical component of the position of the satellite, dy:
dy = V0y*t + (1/2)*a*t^2

We know that the initial velocity in the y-direction is 0, since the satellite is already at its highest point in its orbit. We also know that the acceleration due to gravity, a, is -9.8 m/s^2 (assuming Earth's gravity). So, we can rewrite the equation as:
dy = (1/2)(-9.8 m/s^2)*t^2

Now, we need to find the time, t, it takes for the satellite to travel 162 km. We can use the formula Vx=dx/t to solve for t:
t = dx/Vx

We know that Vx = Vo(cos) angle, so we can substitute that in:
t = dx/(Vo(cos) angle)

We also know that the distance, dx, is the horizontal component of the position of the satellite, so we can rewrite this as:
t = (162 km)/V0x

Now, we can plug this value for t into our equation for dy:
dy = (1/2)(-9.8 m/s^2)*[(162 km)/V0x]^2

We need to convert km to m and angle to radians for the equation to work, so:
dy = (1/2)(-9.8 m/s^2)*[(162,000 m)/(V0x(cos)(62.3 degrees))]^2

We can now solve for dy:
dy = -12,553,500/V0x^2

Next, we can use the equation Vy+V0y/2 = dy/t to solve for V0x:
V0x = (Vy+V0y/2)*t/dy

We know that Vy = V0y = 0, so this simplifies to:
V0x = (0