Post a New Question

college physics

posted by on .

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 7.00s later. What is acceleration of rocket.

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 - ,

    The 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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question