Suppose a 13.0 kg fireworks shell is shot into the air with an initial velocity

Question

Suppose a 13.0 kg fireworks shell is shot into the air with an initial velocity of 64.0 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 from the starting point does the 9.00 kg piece hit the ground? __________m

(b) Calculate the velocity of the 4.00 kg piece just after the separation. __________m/s

(c) At what horizontal distance from the starting point does the 4.00 kg piece hit the ground? ___________m

Explanation / Answer

a)

here

horizontal velocity = 64 * cos(80) =11.1 m/s

vertical = 64 * sin80 = 63.02 m/s

the maximum height is h = v(verticle)^2 / (2g)

h = 63.02^2 / (2*9.8) = 202.6 m

the time to maximum height t = v/ g

t = 63.02 / 9.8 = 6.43 sec

then

the horizontal distance at time of explosion is = t * v(horizontal) = 6.43 * 11.1 = 71.37 m

the 9 kg part have zero velocity and drops vertically down so horizontal distance which is covered is d = 71.37 m

b)

the momentum p1 = m * v(horizontal)^2 = 13 * 11.1 * 11.1 = 1601.73 kg m/s

for 9 kg the momentum is p2 = 9 * 11.1 * 11.1 = 1108.89 kgm/s

the momentum of 4kg particle is = 1601.73 - 1108.89 = 492.84 kgm/s

its velocity v = sqrt(492.84 / 4) + 11.1 = 22.2 m/s

c)

distance = srqt(2*h/g) * v

distance = sqrt( 2 * 202.6 / 9.8) * 22.2 = 142.74 m

the total horizontal distance is = 142.75 + 71.37 = 214.12 m

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