The shell is fired from the ground with an initial velocity u and incination x degree to the horizontal.This is a problem related with projectile motion of the shell. The motion of the shell is detrmined by its initial velocity, the angle of inclination to the horizontal and the acceleration due to gravity at any time of its motion.
The shell has the horizontal and vertical components of velocities at any time: ucosx and usinx -gt, respectively, which are useful to determine the height , vertical and horizontal displacements and the time to reach a postion in the path of the shell.
The vertical displacement s at any time t is given by:
s= (usinx)t-(1/2)gt^2---------------(1), where u is the initial velocity , g is the acceleration due to gravity and x is the direction of angle the shell is fired above horizon.
When the shell hits the ground s=0. Given u=1.7km/s =1700m/s
and x=63 deg and g=9.81m/s^2. Substituting these values in (1) we get:
0= (1700 sin63)t-(1/2)*9.81*t^2. Solving for t, we get the shell time in mtion before reaching the ground.
t=0 or t=2*1700*(sin63)/9.81=308.81 secs. t=0 applies to starting time of the shell. t=308.81seconds is the time the shell remains in motionbeforeit reaches the ground.
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