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Math Help - Rectilinear Motion using integration

  1. #1
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    Rectilinear Motion using integration

    A projectile is launched vertically upward from ground level with an initial velocity of 112 ft/s.
    A. Find the velocity at t= 3s and t=5s
    B. How high will the projectile rise
    C. Find the speed of the projectile when it hits the ground.
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  2. #2
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    Thanks so much in advance. I'm just going over some review problems and I'm a little stuck on this problem.
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  3. #3
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    What have you tried? Since this is going straight up (and then back down, of course) its acceleration is that of gravity, -g. The acceleration is a constant, -g so its speed, at time t is the integral of that. And the distance will be the integral of the speed. The only integral formula you need is \int t^n= \frac{1}{n+1}t^{n+1}+ C.

    Now try to finish it.
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  4. #4
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    Quote Originally Posted by calculusisnotfun View Post
    A projectile is launched vertically upward from ground level with an initial velocity of 112 ft/s.
    A. Find the velocity at t= 3s and t=5s
    B. How high will the projectile rise
    C. Find the speed of the projectile when it hits the ground.

    The equations you have to use are :

    1)v=u-gt ,where v is the final speed u the initial and g the constant accelaration of bodies falling in empty space

    2) h =  ut-\frac{1}{2}gt^2 ,where h is the height t is the time

    So in (1) put t=3s and t= 5s to find the speed

    To find out how high the projectile will rise ,put at the equation(1) v=0 and :

    t= u/g .Now put that into (2) and we get :h = \frac{u^2}{2g}

    Note at the point where the projectile reaches its highest point its v =0

    I will let you do part c
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  5. #5
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    Using integration we have : ( Bold letters denote vectors)

    dv/dt = g.But g =-zg ,where z is the unit vector along the z axis
    And since v = zv we have:

    z \frac {dv}{dt} = -zg and

    dv = -gdt . Integrating the left side from v=u to v=v and the right side from t=0 to t=t we get:

    v=u-gt.

    Also we have :

    dh/dt= v. But h=zh and v=zv.

    Hence zdh = z(u-gt)dt => dh =(u-gt)dt.

    Now integrating the left side from h=0 to h=h and the right side from t=0 to t=t we have:

    h =ut- \frac{1}{2}gt^2
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