You have been asked to design/develop a software tool to simulate a clay target

Question

You have been asked to design/develop a software tool to simulate a clay target shooting (i.e.
determination of the skeet%u2019s trajectory). Create a C++ programming code for the following
case study: A 0.25-kg skeet (clay target) is fired at an angle of 30o
to the horizon with a speed
of25 / m s, as shown in the following image. When it reaches the maximum height, it is hit from below by 15-g pellet travelling vertically
upward at a speed of 200 m/s. The pellet is embedded in the skeet.
(a) How much higher the skeet go up (8 points)?
(b) How much extra distance, %u0394x, does the skeet travel because of the collision (7
points)?
(c) If pellet%u2019s weight is 30-g, calculate how this mass change affects the magnitude of
extra distance %u0394x in section (b)(10 points);
The programming code (*.cpp file) and results from the code should be provided / shown in
the submitted report. Only the C++ implementation of this topic will be accepted

Explanation / Answer

k, you know that when an object reaches its maximum altitude then the vertical velocity is 0 m/s, but it still has a horizontal constant velocity.

Let's find the maximum altitude :

Decomposing the velocity :

Vx = 25*cos(30) = 12.5sqrt(3) m/s

Vy = 25*sin(30) = 12.5 m/s

Using Vy, we will find the maximum altitude :

Knowing that the vertical speed at it is 0 ms :

Vf^2 = Vo^2 - 2gH

Considering g = 10 m/s^2

0 = 12.5^2 - 20H

H = 7.8 meters

Let's find the time, the skeet took :

Vf = Vo -gt

0 = 12.5 -10t

t = 1.25

The horizontal distance in this time : X = 1.25*12.5sqrt(3) = 27 meters.

Now, in that moment is gonna bt hit by a pellet, so let's use the conservation of linear momentum, but only on the vertical.

In other words :

Initial momentum in Y = Final momentum in Y

0.25*0 + 15/1000*200 = 0.265*V'y

V'y = 11.3 m/s

V', is the velocity, that hte skeet and the pellet will have in Y, so the velocity will be : 11.3 m/s in Y, and to find the velocity in X :

0.25*12.5sqrt(3) + 15/1000*0 = 0.265V'x

V'x = 20.4 m/s

New velocity : Vy' = 11.3 ; Vx = 20.4 m/s

Let's find how much higher did the skeet went up :

Initial velocity : 11.3 m/s

Final velocity must be : 0 m/s

g = 10 m/s^2

0 = 11.3^2 - 2*10*H'

H' = 6.4 meters

Let's find the time the skeet took :

0 = 11.3 -10t'

t' = 1.13

The distance : X' = 1.13*12.5sqrt(3) = 24.5 meters

So the skeet went up, 6.4 meters more.

Now, the skeet is : 6.4 +7.8 = 14.2 meters over the ground, so let's find the time, the skeet is gonna take to hit the ground again :

initial velocity = 0m/s

g = 10 m/s^2

H = 14.2 m

Using : H = Vo*t - g*t^2 / 2

14.2 = 1/2*10*t^2

t = 1.68 seconds

distance = 1.68*12.5sqrt(3) = 36.3

Total distance = X + X' + distance : 27 +24.5 +36.3 = 87.8 meters.

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