An attack helicopter is equipped with a 20
mm cannon that fires 116 g shells in the
forward direction with a muzzle speed of
819 m/s. The fully loaded helicopter has a
mass of 4870 kg. A burst of 96.2 shells is fired
in a 4.51 s interval.
What is the resulting average force on the
helicopter?
Answer in units of N
By what amount is its forward speed reduced?
Answer in units of m/s
How do i find the amount that the forward speed is reduced once i find the resulting average force on the helicopter?(2026.47113082N)
1 answer
The amount that the forward speed is reduced can be calculated using the equation F = ma, where F is the force, m is the mass of the helicopter, and a is the acceleration. Rearranging this equation, we get a = F/m. Plugging in the values for F (2026.47113082N) and m (4870 kg), we get a = 0.415 m/s^2. This is the acceleration of the helicopter due to the force of the cannon fire. To find the amount that the forward speed is reduced, we can use the equation vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time interval. Plugging in the values for vi (819 m/s), a (0.415 m/s^2), and t (4.51 s), we get vf = 819 m/s - 1.845 m/s = 817.155 m/s. Therefore, the forward speed of the helicopter is reduced by 1.845 m/s.