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

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 theta= 67 degrees, the crew fires the shell at a muzzle velocity of 131 ft per second. How far down the hill does the shell strike if the hill subtends an angle 32 degrees from horizontal? (Ignore air friction.) How long will the mortar shell remain in the air? How fast will the shell be traveling when it hits the ground?

Explanation / Answer

EditedVersion

The answer is right, i have devouted lots of time to slove this question, there may be some calculation mistake but the method is right we are changing the coordinate system

taking along slope as x axis and perpendicual to it as y axis accodingly thee value of g changes.

we are changing the co-ordinate system

and make it along the hill ;

So,

vertical acceleration = -g*sin 32 = 17.3489 ft/sec
Horizontal acceleration = g*cos 32 = 27.765 ft/sec
0 = 131 *sin(67-32) - g*sin 32 *t

0 = 75.1385 - 17.3489*t
t = 4.33102 s
total time =2*t = 8.6620 sec
x = 131*cos(67-32)*8.6620 +0.5*27.765*T^2 = 929.509 +

x = 1971.11 ft
a)x = 1971.11 ft (ans to a)
b) T = 8.6620 s
c) vy = 131 *sin(35) - 17.3489*8.6620 = -75.13 ft/s
vx = 131 *cos(35) + 27.765 *8.6620 = 347.809ft/s
speed = 355.83 ft/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.