SOLUTION: An arrow is shot straight up in the air with an initial speed of 220 ft/s. If on striking the ground it embeds itself 2.00 in into the ground, find the magnitude of the acceleratio
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Question 1047598: An arrow is shot straight up in the air with an initial speed of 220 ft/s. If on striking the ground it embeds itself 2.00 in into the ground, find the magnitude of the acceleration (assumed constant) required to stop the arrow, in units of feet/second^2.
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
The arrow is fired 220 ft/s straight up, reaches a maximum height, then falls straight down. The final velocity when it reaches the ground is then used to find the acceleration needed to go from that velocity to zero, over a distance of 2.00 inches, or 2/12 = 1/6 ft.
:
since the arrow is fired straight up, the y component is sin(90) * 220 = 220
:
when the arrow reaches max height its velocity is 0, therefore
:
0 = 220 + (-32t)
:
note that 32 is acceleration of gravity acting downward on the arrow
:
t = 220 / 32 = 6.975 seconds
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max height of arrow = 220 * 6.975 = 1512.5 feet
:
at max height the arrow has potential energy that equals its kinetic energy at ground 0
:
mgh = (1/2)mv^2
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32(1512.5) = (1/2)v^2
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v^2 = 96800
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v = 311.1 ft/sec
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now use the equation
:
v(f)^2 = v(0)^2 + 2as
:
0 = 311.1^2 + 2a(1/6)
:
(1/3)*(a) = -96800
:
a = -290400
:
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magnitude of acceleration to stop arrow is 290400 ft/sec^2
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