A 3.15-kg projectile is fired with an initial speed of 119 m/s at an angle of 31

Question

A 3.15-kg projectile is fired with an initial speed of 119 m/s at an angle of 31 with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 0.97 kg and 2.18 kg. At 3.57 s after the explosion, the 2.18-kg fragment lands on the ground directly below the point of explosion.

(b) Find the distance between the point of firing and the point at which the 0.97-kg fragment strikes the ground.
____km   

(c) Determine the energy released in the explosion.
____kJ

A 3.15-kg projectile is fired with an initial speed of 119 m/s at an angle of 31 with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 0.97 kg and 2.18 kg. At 3.57 s after the explosion, the 2.18-kg fragment lands on the ground directly below the point of explosion. (a) Determine the velocity of the 0.97-kg fragment immediately after the explosion. j hat (b) Find the distance between the point of firing and the point at which the 0.97-kg fragment strikes the ground. ____km (c) Determine the energy released in the explosion. ____kJ i hat + ____m/s vector v 1 = ____m/s

Explanation / Answer

Vx = V*cos31 = 102 m/s...........Vy =V*sin31 = 61.28 m/s

H = Vy^2/2g = 191.6 m/s

for 2.18 kg

-H = V1y*T - 0.5*g*T^2

-191.6 = v1y*3.57 - (0.5*9.8*3.57*3.57)

V1y = -36.17 m/s

at the top

(m1+m2)*Vx = m1*V1x + m2*V2x

3.15*102 = 0 + 0.97*V2x

V2x = 331.23 m/s

along y

(m1+m2)*0 = -m1*36.17 + 0.97*V2y

V2y = 37.29 m/s

V = 331.23i + 37.29 j

V = 333.32 m/s

K1 = 0.5*3.15*102*102 = 16386.3 J

K2 = (0.5*2.18*36.17*36.17) +(0.5*0.97*333.32*333.32) = 55310.590965 J

K2 - K1 = 38924.290965 J = 38.9 KJ

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