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DYNAMICS PROBLEMS

Given that the total launch speed with respect to the ground is
U_ground = U_climb + U_launch = 200 + 50 = 250km ⁄ h = 69.4m ⁄ s

Since the angle of launch is 45^○, vertical launch velocity of the bomb with respect to the ground is v_o = U_groundsin45.
(7) v_o = 69.4sin(45) = 69.4 ⁄ √(2) = 49.1m ⁄ s upwards

Horizontal launch velocity of the bomb with respect to the ground is u_o = U_groundcos45.
(8) u_o = 69.4cos(45) = 69.4 ⁄ √(2) = 49.1m ⁄ s forward
Since the bomb is being launched at an angle of 45 degree upward, it experiences a deceleration in the y -direction. Analyzing the FBD (free body diagram) for forces in the vertical direction on the bomb,
F = ma_y = − W = − mg
where a_y represents the acceleration of the bomb in the upward direction. Thus by substituting a_y = − g, the velocity in the y-direction becomes:
(9) ^v⌠⌡_{v_o}dv = ^t⌠⌡₀a_ydt

⇒ v(t) = v_o − gt
Now, distance travelled by the bomb in the y-direction when it is dropped from a height H is given by:
(10) ^y⌠⌡_Hdh = ^t⌠⌡₀vdt

⇒ y(t) − H = ^t⌠⌡₀(v₀ − gt)dt

⇒ y(t) = H + v_ot − (1)/(2)gt²
where y is the instantaneous height of the bomb with respect to the ground. Total time of flight of the bomb (time taken for the bomb to fall to ground) is found by substituting y(t) = 0 in the above equation. Thus,
H + v_ot − (1)/(2)gt² = 0
Upon solving this quadratic expression for the time of flight (t):
t_flight = (v_o)/(g) + √(⎛⎝(v_o)/(g)⎞⎠² + H)
where v_o = U_groundsinθ = 49.1m ⁄ s, H = 500m and assume acceleration due to gravity (g = 9.8m ⁄ s). Substituting these values, results in t_flight = 27.93 s.

The equation for obtaining the Range (horizontal distance covered by the bomb on the ground) is given by:

(11) R = u_ot_flight

(12) R = u_ot = 49.1 × 27.93

Therefore horizontal range (R) is given by R = 1371.36m

References

http://en.wikipedia.org/wiki/Anti-lockbrakingsystem

http://www.smartcockpit.com/data/pdfs/flightops/flyingtechnique/SlipperyRunways.pdf

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