An aircraft with a mass of 5000 kg takes off from sea level with a forward speed of 50 m/s and starts to climb with a climb angle of 15 degree. What is the rate of climb and excess thrust available at the start of the climb?
An aircraft with a mass of 5000 kg takes off from sea level with a forward speed of 50 m/s and starts to climb with a climb angle of 15 \(^{\circ}\). What is the rate of climb and excess thrust available at the start of the climb?
The aircraft is climbing with an climb angle of 15\(^{\circ} \).Here forward speed is 50 m/s.From the figure we can see that rate of climb will be given as \[Vsin\theta
\\=50sin15^{\circ}
\\=12.94\;m/s\]
Excess thrust is given as \(T-D=Wsin\theta\)
\[=5000\times 9.81 \times sin15^{\circ}\]
\[\\=12695.1 N\]
