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 theta = 58.0 degree (as shown), the crew fires the shell at a muzzle velocity of 176 feet per second. How far down the hill does the shell strike if the hill subtends an angle Phi = 34.0 degree from the 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

V = 53.6 m/sec
sin 34° = 0.559
cos 34° = 0.829
-y = 53.6*sin40°*t-4.9t^2 .... -y = 34.45t-4.9t^2 (1)
0.829y/0.559 = 53.6*cos58°*t .... 1.483y = 28.40t ..... -y = -19.15t (2)
equating both equations (1) and (2)
-19.15t = 34.45t-4.9t^2
4.9t^2 = 53.6t
t = 53.6/4.9 = 10.93 sec
d = 28.40*10.93 = 310.412 m
h = 34.45*10.93-4,9*10.93^2 = -208.83 m
1/2mVf^2 = mgh+1/2mV^2
Vf = 2gh+V^2 = 19.6*208.83+53.6^2 = 83.46 m/sec approx.

How long will the mortar shell remain in the air?
t = 53.6/4.9 = 10.93 sec

How fast will the shell be traveling when it hits the ground?
Vf = 2gh+V^2 = 19.6*208.83+53.6^2 = 83.46 m/sec appro

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