Question

. A mortar is a simple projectile which is launched at a fixed velocity v0, but the angle θ at which it is launched can be varied. An artillery officer finds that he can hit a target a maximum distance of 2.25 kilometers away when launched on the flat horizontal practice range. The officer is later charged with using the same mortar to strike an enemy ammunition depot before it can be moved. He determines that the depot is a distance of 1 kilometer away, but lies 10 meters below his current location. At what angle should he aim the mortar so that he strikes the target in the minimum time possible? Again, use g = 10 m/s2 here and be sure to draw diagrams of the problem. (12.56◦ above the horizontal)

Answer 1

maximum range = v_o²/g = 2250 => v_o = 150 m/s

now to hit target, R = 1000 m

H = -10 m

let A be the angle above horizontal

=> 1000 = (150cosA)t

=> -10 = (150sinA)t - 0.5gt²

=>> -10 = (150sinA)(1000/150cosA) - 5(1000/150cosA)²

=> -10 = 1000tanA - 2000/9cos²A

=> (200sec²A)/9 - 100tanA - 1 = 0

=> 200(tan²A + 1)/9 - 100tanA - 1 = 0

=> 200tan²A - 900tanA + 191 = 0

=> tanA = 4.2767, 0.2233

for minimum time, tanA = 0.2233

=> A = 12.58^o