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January 31, 2015

January 31, 2015

Posted by **Drew** on Sunday, September 6, 2009 at 10:46pm.

I believe if you use x(final)= x(initial +v(initial)deltaT+ 1/2adeltatsquared you get something like x=0+0+1/2(-9.8)(4)^2 and that will give distance of the rocket when the bolt fell off. then use the same equation only plug in 240 for x final and plug in 4 for T and solve for a, but when you do that you get 30.01 and that is incorrect for some reason.

- college physics -
**MathMate**, Sunday, September 6, 2009 at 11:34pmThe rocket starts from ground, where h=0.

Let

a=upward acceleration of rocket (m/s/s)

t1=4 s

t2=7 s from time when holt falls off

u=velocity of rocket (and bolt) after t1 seconds

= a*t1 m/s

=4a m/s

s1=distance from ground after t1 seconds

=0*t1 + (1/2)a(t1)²

=(1/2)a(t1)²

=8a m

The height h of the bolt from the ground after it falls off

h=u*t2+(1/2)(-g)(t2)²

=4a*t2 + (1/2)(-9.81)(7²)

=28a + 240.345

Therefore

-8a = 28a + 240.345

a=6.676 m/s/s

Check:

u=4a=26.70 m/s

s1=8a=53.4 m

h=u*t2+(1/2)(-g)(t2)²

=26.70*7 + (1/2)(-9.81)(49)

= -53.4 m from the point bolt dropped off = u OK

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