An airplane whose speed is 150km/hr climbs from a runway at an angle of 20° above the horizontal. What is its altitude two minutes after takeoff? How many kilometers does it travel in a horizontal direction during this two minute period?
My work:
150cos(20) = 140.9
150sin(20) = 51.3
Unfortunately, that wasn't the right answer. Please help!
You work isn't correct.
2 min= 1/30 hr
horizontal distance=150km/hr*1/30 hr*cos20
vertical distance=150km/hr*1/30 hr*sin20
To solve this problem, you need to use the equations of motion and trigonometry to find the altitude and horizontal distance traveled by the airplane.
First, let's find the altitude after two minutes:
We know that the airplane's speed is 150 km/hr, which means it covers a distance of 150 km in one hour. Since we are interested in the time frame of two minutes, we need to convert 150 km/hr to km/minute.
To do this, divide the speed by 60, as there are 60 minutes in an hour:
150 km/hr ÷ 60 = 2.5 km/minute
Now, we can calculate the altitude using trigonometry.
The altitude is given by the equation:
altitude = distance × sin(angle)
So, the altitude after two minutes is:
altitude = 2.5 km/min × sin(20°) = 0.8585 km
Therefore, the altitude after two minutes is approximately 0.8585 km.
Now let's find the horizontal distance traveled during this two-minute period:
The horizontal distance is given by the equation:
horizontal distance = distance × cos(angle)
So, the horizontal distance after two minutes is:
horizontal distance = 2.5 km/min × cos(20°) = 2.3544 km
Therefore, the horizontal distance traveled during this two-minute period is approximately 2.3544 km.
I apologize if my previous response didn't match the correct answer. Please let me know if there's anything else I can assist you with.