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In these "projectile problems" the idea is to break the problem into two separate motions. In the Vertical direction, the object moves according to "constant acceleration" (of gravity) equations. In the Horizontal direction the object moves according to "constant velocity" equations.

=>It turns out that in this particular question only the Vertical motion need be considered...more generally the question includes "how far horizontally the shell travels which would require use of horizontal component of velocity and constant velocity equation)

An initial velocity of 1800 m/s at 38.6° means:

Vertical velocity component = v0y = 1800 sin 38.6 = 1123 m/s

Horiz. velocity component = vox = 1800 cos 38.6 = 1407 m/s

t = total time of flight of shell

Since at the end of flight the shell returns to the ground its vertical displacement at time = t is 0.

Vertical displacement = (initial vertical velocity)(t) + 1/2 gt²

{initial vert. velocity is up and "g" acts down so use neg sign}

0 = v0y t - 1/2 gt²

0 = 1123t - 4.9t²

1123t = 4.9t²

1123 = 4.9 t

t = 1123/4.9 = 229 seconds (ANS)

For a projectile, the total flight time is given by the formula

T = 2Vo(sin Θ)/g

where

Vo = initial velocity = 1800 m/sec. (given)

Θ = angle of projection = 38.6 deg (given)

g = acceleration due to gravity = 9.8 m/sec^2

Substituting values,

T = 2(1800)(sin 38.6)/9.8

T = 229.2 seconds = 3.82 min

Hope this helps.

The vertical area of velocity is sine (seventy 3.8°). The horizontal area of velocity is cosine (seventy 3.8°) Ke = Pe, mgh = ½mv² , gh = ½v², h = v²/2g the optimal altitude is = [[sine (seventy 3.8°)×a million,570m/s]² ÷ 2 × 9.8m/s² s = ½at² , t = ?(2s/a) making use of the optimal altitude, the time of flight is = ?[ 2 × (max top)/9.8m/s²] × 2 Distance = velocity × time the gap traveled horizontally is t (from above) × cosine (seventy 3.8°) × a million,570 m/s

need the weight of the shell