A 3.03-kg projectile is fired with an initial speed of 129 m/s at an angle of 28

Question

A 3.03-kg projectile is fired with an initial speed of 129 m/s at an angle of 28° with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 1.11 kg and 1.92 kg. At 3.93 s after the explosion, the 1.92-kg fragment lands on the ground directly below the point of explosion. (a) Determine the velocity of the 1.11-kg fragment immediately after the explosion. vector v1 = m/s i hat + m/s j (b) Find the distance between the point of firing and the point at which the 1.11-kg fragment strikes the ground. km (c) Determine the energy released in the explosion. kJ

Explanation / Answer

m = 3.03 Kg
Vx = 129 * cos(28) = 113.9 m/s
Vy = 129 * sin(28) = 60.56 m/s

At top of it's trajectory, Vy = 0
v^2 = u^2 - 2*a*s
0 = 60.56^2 - 2*9.8*s
s = 187.1 m

t = u/g = 60.56/9.8
sx = 113.9 * ( 60.56/9.8) = 703.8 m
Horizontal distance travelled, = 703.8 m

Now,
t = 3.93 s
Vx = 703.8/3.93 = 179.1 m/s
Vy = 187.1/3.93 = 47.6 m/s

Using Momentum Conservation,
In X DIRECTION,
3.03 * 113.9 = - 1.92 * 179.1 + 1.11 * V
Vx = 1.12 m/s

In Y DIRECTION,
0 = -1.92 * 47.6 + 1.11 * V
Vy = 82.33 m/s

v1 = 620.7 m/s i^ + 82.33 m/s j^

(b)
Distance travelled by 1.11 Kg Fragment,
-187.1 = 82.33*t - 1/2*9.8*t^2
t = 18.83 s
Total Horizontal distance covered, = 18.83 * 620.7 = 11687.7 m

Distance between the point of firing and the point at which the 1.11-kg fragment strikes the ground = 703.8 + 11687.7 = 12.4 km

(C)
Energy Releases,
Initial K.E - Final K.E
1/2*3.03*113.9^2 - 1/2*1.92 * (179.1^2 + 47.6^2) - 1/2 * 1.11 * (620.7^2 + 82.33^2 )
= 230.9 kJ

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