A mortar* crew is positioned near the top of a steep hill. Enemy forces are char

Question

A mortar* crew is positioned near the top of a steep hill. Enemy forces are charging up the hill and it is necessary for the crew to spring into action. Angling the mortar at an angle of ? = 61.0 degrees, the crew fires the shell at a muzzle velocity of 249 feet per second. a)How far down the hill does the shell strike if the hill subtends an angle ? = 36.0 degrees from the horizontal? (Ignore air friction.) b)How long will the mortar shell remain in the air? c)How fast will the shell be traveling when it hits the ground? * The mortar is like a small cannon that launches shells at steep angles.

Explanation / Answer

we will change the co-ordinate system and make it along the hill ; So vertical acceleration = g*sin 36 Horizontal acceleration = g*cos 36 0 = 249 *sin(61-36) - g*sin 36 *t t = 5.5644 s total time =2*t = 11.1289 s x = v*cos(61-36)*T -0.5*32.174*cos 36 T^2 a)x = 899.566 ft (ans to a) b) T = 11.13 s c) vy = 249 *sin(61-36) - g*sin 36 *T = -108.288 ft/s vx = 249 *cos(61-36) - g*cos 36 *T = -68.174 ft/s speed = 127.936 ft/s

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