Suppose a 12.0 kg fireworks shell is shot into the air with an initial velocity

Question

Suppose a 12.0 kg fireworks shell is shot into the air with an initial velocity of 71.0 m/s at an angle of 80.0° above the horizontal. At the highest point of its trajectory, a small explosive charge separates it into two pieces, neither of which ignite (two duds). One 9.00 kg piece falls straight down, having zero velocity just after the explosion. Neglect air resistance (a poor approximation, but do it anyway).

a) At what horizontal distance (in m) from the starting point does the 9.00 kg piece hit the ground?

(b) Calculate the velocity (in m/s) of the 3.00 kg piece just after the separation.

(c) At what horizontal distance (in m) from the starting point does the 3.00 kg piece hit the ground?

Explanation / Answer

a) Required Horizontal distnce is Horizontal component of velocity * time of ascent

R = u*cos(80)*u*sin(80)/g = 71*cos(80)*71*sin(80)/9.8 = 87.96 m

b) using law of conservation of momentum

along X-axis

12*u*cos(80) = 9*v1*cos(80) + 3*v2*cos(80)

12*71*cos(80) = 9*v1*cos(80) + 3*v2*cos(80)

147.94 = 1.56*v1 + 0.52*v2

along Y-axis

12*71*sin(80) = 9*0 + 3*v2*sin(80)

839 = 2.95*v2

v2 = 284.4 m/sec

c) required horizontal distance is 87.96 + (v2*sqrt(2h/g))

h = u^2*sin^2(80)/(2*g) = 71^2*sin^2(80)/(2*9.8) = 249.43 m

then R = 87.96+ (284.4*sqrt(2*249.43/9.8)) = 2117 m

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