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An object is thrown horizontally with a velocity of 30 m/s from the top of a tower. It undergoes a constant downward acceleration of 10 m/s2. The magnitude of its instantaneous velocity after 4.0 sec, in meters per second, is:

To approach this question I first thought to myself that the velocity in the y-component after 4s is going to equal 10+2(10)+3(10)+4(10); 100m/s. The x velocity will remain constant. Thus the velocity at t=4 would be the resultant vector of 100m/s in the y and 30m/s in the x, which equals 104.4m/s. I am wondering where I am going wrong with my reasoning here?

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1 Answer

up vote 0 down vote accepted

In $y$ direction you have accelerated movement with constant acceleration, thus

$$v_y = v_{y0} - g t$$

and after putting initial conditions

$$|v_y| = g t$$

I have no idea whatsoever what did you want to do with your calculation.

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Ahh, I was assuming that the rate of change of acceleration was 10m/s. But its constant acceleration - so each second travels = another 10m/s added to the velocity. –  Kurt Apr 15 '12 at 19:58
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That's true, and it will be 20m/s at t=2, 30m/s at t=3 and 40m/s at t=4... –  Pygmalion Apr 15 '12 at 19:59
    
@Kurt Check mark the answer, then.. –  Sachin Shekhar Jul 14 '12 at 22:09
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protected by Qmechanic Oct 28 '13 at 14:15

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