Respuesta :
Answer:
The location is at (3.535, 1.162) m
Solution:
As per the question:
Mass of the projectile, m = 9.6 kg
Initial velocity, v = 12.4 m/s
Angle, [tex]\theta = 54^{\circ}[/tex]
Mass of one fragment, m = 6.5 kg
Time taken by the fragment, t = 1.42 s
Height of the fragment, y = 5.9 m
Horizontal distance, x = 13.6 m
Now,
To determine the location of the second fragment:
Horizontal Range, [tex]R = \frac{v^{2}sin2\theta}{g}[/tex]
[tex]R = \frac{12.4^{2}sin2(54)}{9.8} = 14.92\ m[/tex]
Time of flight, t' = [tex]\frac{2vsin\theta}{g} = \frac{2\times 12.4sin108}{9.8}= 2.406\ s[/tex]
Now, for the fragments:
Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg
Distance traveled horizontally:
[tex]s_{x} = vcos\theta = 12.4cos54^{\circ}\times 1.42 = 10.35\ m[/tex]
Distance traveled vertically:
[tex]s_{y} = vcos\theta - \frac{1}{2}gt^{2}[/tex]
[tex]s_{y} = 12.4sin54^{\circ}\times 1.42 - \frac{1}{2}\times 9.8\times 1.42^{2} = 14.25 - 9.88 = 4.37\ m[/tex]
Now,
[tex]s_{x} = \frac{mx + m'x'}{M}[/tex]
[tex]10.35= \frac{6.5\times 13.6 + 3.1x'}{9.6}[/tex]
x' = 3.535 m
Similarly,
[tex]s_{y} = \frac{my + m'y'}{M}[/tex]
[tex]4.37= \frac{6.5\times 5.9 + 3.1y'}{9.6} = 1.162\ m[/tex]
The location of the other fragment is at (3.535, 1.162)
