Question

Frequently, a weapon must be fired at a target that is closer than the weapon's maximum range. To hit such a target, a weapon has two possible launch angles (figure A): one higher than 45= (x) and one lower than 45- (C.). Although the displacement of the projectile is the same for the two angles, a projectile launched at y has a longer flight time and a higher peak position than one launched at 8. Usually, some tactical situation makes one angle preferable to the other. For example, if the projectile must go over some nearby object such as a grove of trees, the higher angle may be desirable. A shorter flight time and therefore 0, are preferable if the target is mobile, In practice, many weapons are designed to operate either at angles lower than 45º or at angles higher than 45º, but not both. Tanks, for example, often must face mobile targets; to minimize the time the target has to move, tanks fire at low angles. Grenades, on the other hand, are launched at high angles because a soldier launching a grenade is often close to the target, but has no armor plating for protection. The high launch angle allows the soldier to stay out of sight by hiding behind some obstacle, and the longer flight time may make it possible for the soldier to move farther from the exploding grenade. Imagine an unusual scenario in which a large gun mounted on a vehicle is required to hit an explosives factory (figure B). A huge explosion expected, and there must be time for the gunner to retreat. A grove of trees provides cover. The maximum range of the gun is 17.1 km, and the maximum speed of the vehicle is 66.8 km/h. (a) What is the muzzle speed Vo? (Muzzle speed is the speed at which the projectile leaves the barrel of the gun. Enter your answer to at least three significant digits.) m/s (6) The target is 5.5 km away. Find the low angle 8 and the high angle 8,, at which the gunner may alm so as to hit the target. 0 = counterclockwise from the +x axis = - counterclockwise from the +x axls (c) Find the time the projectile takes to hit the target for both angles. t = s TH= s (d) Assume the vehicle retreats at its maximum speed (66.8 km/h) to be as far from the ensuing explosion as possible. How far is it from the factory at the time of the explosion for each launch angle? XH = m

Answer 1

The maxi num Range occuurs at an angle of 45 and it is given by formal a Rmax=628m2o , U- opeedot brojcetin g angle of projection 450 17-1=Us ² X Sngo maxinuum fange = 17.1km 9.8 =171xlorom Vol. 9-8 X171 X1000 a) Uo=4.094 10² mp. 409 mip Rmax- 409.4 mio. b) Now Range which we R = 5.5kmi have to covert R = (109.4)²x5nza R v. 2 smzo 9.8 g Smzo = 9.847 5.5X 16000 98x1718 2000 528 =18.760 0 = 9.38° OL 9.38° And OH= 90-938°= 8064

C2 Time of flight - qusno g = te=quo Smal g ItL= 13.6L sech 2X409.4X5on 9.30 9.8 fom (180-61) = th= quo SongH 2x409.4% 9.2 g tH = 82-5oee Now the relicle can retreat at its naxinum & peed of 66.8 km = 18.56 m2 =0 er Distance speed & time: XL = Q* tl = 252-54 my C minimuen distance) maximum distance TH= v*th XH = 18-56x82-5=1530.83