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.
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.
2 answers
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.
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.