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? Find the altitude and horizontal distance.
150cos(20) = 140.9
150sin(20) = 51
Please help.
V = 150km/h[20o]
Vx = 150*cos20 = 141 km/h
Vy = 150*sin20 = 51 km/h.
a. T = 2min. * 1h/60min = 2/60=1/30 h.
Altitude = Vy*T = 51 * 1/30 = 1.71 km.
b. Dx = Vx*T = 141 * 1/30 = 4.7 km.
V = 150km/h[20o]
Vx = 150*cos20 = 141 km/h
Vy = 150*sin20 = 51 km/h.
a. T = 2min. * 1h/60min = 2/60=1/30 h.
Altitude = Vy*T = 51 * 1/30 = 1.71 km.
b. Dx = Vx*T = 141 * 1/30 = 4.7 km.
c. Total Distance = V*T = 150 * 1/30 =
5 km/h.
To find the altitude and horizontal distance traveled by the airplane, we can use the given information about its speed and angle of climb.
First, let's find the altitude of the airplane after two minutes. Since we know that the speed of the airplane is 150 km/hr, we can convert this to km/min by dividing it by 60 (since there are 60 minutes in an hour). So, the speed of the airplane is 150 km/hr / 60 min/hr = 2.5 km/min.
To find the altitude, we need to use the trigonometric relationship between the angle of climb and the altitude. The altitude (h) is given by the equation:
h = speed * sin(angle)
Substituting the values we have:
h = 2.5 km/min * sin(20°)
Using a calculator, we find:
h ≈ 0.86 km
Therefore, the altitude of the airplane two minutes after takeoff is approximately 0.86 km.
Next, let's determine the horizontal distance traveled by the airplane during this two-minute period. The horizontal distance is given by the equation:
distance = speed * cos(angle) * time
Substituting the values we have:
distance = 2.5 km/min * cos(20°) * 2 min
Using a calculator, we find:
distance ≈ 4.90 km
Therefore, the horizontal distance traveled by the airplane during this two-minute period is approximately 4.90 km.
To summarize:
- The altitude of the airplane two minutes after takeoff is approximately 0.86 km.
- The horizontal distance traveled by the airplane during this two-minute period is approximately 4.90 km.