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

Question

Map 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 = 58.09(as shown), the crew fires the shell at a muzzle velocity of 187 feet per second. How far down the hill does the shell strike if the hill subtends an angle = 39.00 from the horizontal? (Ignore air friction.) Number How long will the mortar shell remain in the air? Number How fast will the shell be traveling when it hits the ground? Number m/s * The mortar is like a small cannon that launches shells at steep angles.

Explanation / Answer

a)given

vo = 187 ft/s

= 187*0.3048 m/s

= 57 m/s

vox = 57*cos(58) = 30.2 m/s
voy = 57*sin(58) = 48.3 m/s

let t is the time taken for shell to hit the hill.

in x-direction,

t = d*cos(phi)/vox

= d*cos(39)/30.2

in y-direction,

-d*sin(phi) = voy*t - (1/2)*g*t^2

-d*sin(39) = 48.3*d*cos(39)/30.2 - (1/2)*9.8*(d*cos(39)/30.2)^2

==> d = 577 m

B) t = = d*cos(39)/30.2

= 577*cos(39)/30.2

= 14.8 s

C) vx = vox

= 30.2 m/s

vy = voy - g*t

= 48.3 - 9.8*14.8

= -96.7 m/s

so, v = sqrt(vx^2 + vy^2)

= sqrt(30.2^2 + 96.7^2)

= 101.3 m/s

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