# How long is the shell in motion? Answer in units of s.A shell is fired from the ground with an initial speed of 1.70 x 10^3 m/s (approximately five times the speed of sound) at an initial angle of...

How long is the shell in motion? Answer in units of s.

A shell is fired from the ground with an initial speed of 1.70 x 10^3 m/s (approximately five times the speed of sound) at an initial angle of 63.0 degrees to the horizontal. The acceleration of gravity is 9.81 m/s^2.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The shell is fired from the groundÂ  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.

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

The motion of the shell can be separated in two components. One is the horizontal movement at steady velocity. The other is a vertical movement at a velocity which changes under the gravitational force. The acceleration of this change in vertical velocity is equal to 9.81 m/s^2.

Under the influence of vertical velocity combined with acceleration the shell rises in vertical direction till it reaches zero velocity and then starts falling down till it hits the ground. The time taken by the shell during its fall is equal to the the time taken for its rise.

It is assumed that the shell explodes as it hits the ground and therefore does not travel further. Therefore the the shell is in motion during the period of its rise and fall.

Solution.

Given Real firing velocity = 1.7 * 10^3 = 1700 m/s

Angle of firing = 63 degrees to horizontal

Vertical velocity of the shell = Real firing velocity * sin(angle with horizontal

= 1700*(sin63) = 1700*0.891 = 1517.7

The time taken to reach highest point = initial velocity/acceleration

= 1517.7/9.81 = 154.40366 seconds approximately

Time for which the shell is in motion = 2* time of rising to highest point

= 2*154.40366 = 308.8073 s (approximately) Answer: The shell is in motion for 308.8073 s