Consider fireworks launched from the ground. The initial charge fires the firewo

Question

Consider fireworks launched from the ground. The initial charge fires the firework with an initial velocity of 46m/s straight up. When the firework explodes at its maximum height, hot pieces of cardboard are sent flying in all directions. If the minimum safe distance for public viewing is established so no hot pieces of cardboard can reach you, calculate the minimum safe distance if the hot pieces are ejected at 60m/s. Given we are ignoring air resistance and the cardboard pieces are very light, this result is far more than enough.

Explanation / Answer

along vertical

initial velocity voy = 46 m/s

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

at maximum height final velocity vy = 0

displacement y = h

from equation of motion

vy^2 - voy^2 = 2*ay*y

0 - 46^2 = -2*9.8*h

h = voy^2/2g = 46^2/(2*9.8)

after the hot pieces are flied

along vertical

initial velocity voy = 0

acceleration ay = -g

displacement y = -h

y = voy*T + (1/2)*ay*T^2

-46^2/(2*9.8) = 0 -(1/2)*9.8*T^2

T = 46/g = 46/9.8 = 4.69 s

along horizontal

initial velocity vox = 60 m/s

acceleration ax = 0

displacement x = vox*T + (1/2)*ax*T^2

x = 60*4.69

x = 281.4 m <<<<<------ANSWER

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