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 = 53.0o (as shown), the crew fires the shell at a muzzle velocity of 201 feet per second. How far down the hill does the shell strike if the hill subtends an angle = 37.0o 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?

Please keep in mind that the answer needs to be in meters and meters/seconds not in feet.

Explanation / Answer

let;s start with working out the equation for distance travlled by mortar

D = ( V cos 53 t ) + ( V sin 53 t - 1/2 gt^2)

V = 201 ft /seconds= 61.2648 m/s

let mortar tavel a distance of d on the hill

-d sin = Y

dcos = X

Y/ X=- tan = ( V sin 53 t - 1/2 gt^2)/ ( V cos 53 t )

-tan 37 = ( 61.2648 sin 53 - 4.9 t) / ( 61.2648 cos 53)

-tan 37 = tan 53 - 4.9 t/ ( 61.2648 cos 53)

-0.753 = 1.327 - 0.133 t

t = 15..64 seconds apprx

X ( horizontal distance travelled) =  61.2648 cos 53 ( 15.64) = 576.648 m

Y =  ( 61.2648 x 15.64 sin 53 - 4.9 (15.64)^2= -433.347 ( -ve sign indicate it ' stravelling in - y direction)

D = sqroot (X^2 + Y^2) = 721.33 m apprx

c) Vf = sqroot ( vx^2 + vy^2)=

Vx=61.2648 cos 53= 36.87 m/s

Vy = 61.2648 sin 53 - 9;8 (15..64) =-104. 344

V =110.666 m/s appr

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