Part A In a fireworks display, a rocket is launched from the ground with a speed

Question

Part A In a fireworks display, a rocket is launched from the ground with a speed of 27.0 m/s and a direction of 51.0° above the horizontal. During the flight, the rocket explodes into two pieces of equal mass (Figure 1) What horizontal distance from the launch point will the center of mass of the two pieces be after both have landed on the ground? Express your answer with the appropriate units R Value Units Submit My Answers Give Up Figure 1 of 1 Part B If one piece lands a horizontal distance of 26.0 m from the launch point, where does the other piece land? Find the distance from the launch point. Express your answer with the appropriate units After the shell esplodes, the two fragment fellow individoal trajectories ut the ocr of mass (m continues to follow the Shell explodes d-Value UUnits Submit My Answers Give Up Continue

Explanation / Answer

1)

along horizontal
________________

initial velocity v0x = v*cos(theta)

acceleration ax = 0

displacement x = R

from equation of motion

x = v0x*T+ 0.5*ax*T^2

x = v*costheta*T

T = x/(v*cos(theta))......(1)

along vertical
______________

v0y = v*sin(theta)

acceleration ay = -g = -9.8 m/s^2

from equation of motion

y-y0 = vy*T + 0.5*ay*T^2

y-y0 = (v*sin(theta)*x)/(v*cos(theta)) - (0.5*g*x^2)/(v^2*(cos(theta))^2)

y-y0 = x*tan(theta) - ((0.5*g*x^2)/(v^2*(cos(theta))^2))

after falling on ground y - y0 = 0

0 = R*tan51 - (0.5*9.8*R^2/(27*cos51)^2)

R = 72.8 m <<<-----ANSWER

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part B

mass of first piece m1 = m

x1 = 26 m

mass of second piece m2 = m

x2 = d

Xcm = R = 72.8 m

Xcm = ((m1*x1) + (m2*x2))/(m1+m2)

72.8 = ((m*26) + (m2*d))/(m+m)

72.8 = (26+d)/(2)

d = 119.6 m

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