Acapacitor, fully charged by a 3.46v battery, is able to light a small 3.00-w bulb at steady full power for 8.00 s before it quits. what is the capacitance of the capacitor?

First, I'm going to calculate how much energy the little bulb uses to dissipate 3 watts for 8 seconds.

Then, I'm going to calculate what size capacitor stores that amount of energy when charged to 3.46 volts.

3 watts means 3 joules per second

(3 joules/sec) x (8 sec) = 24 joules

Now, energy stored in a capacitor = (1/2) (capacitance) (voltage)²

24 joules = (1/2) (3.46 v)² (capacitance)

Capacitance = (24 J) / (0.5 x 11.9716)

Capacitance =  4.01 farads

Just as I suspected when I read the question . . . the answer is a ginormous,  humongously large capacitance.  Typical capacitors, that you're likely to have in your ' C ' box under your work bench, are all labeled in units of 'microfarads' or 'picofarads'.  Even the huge, low-voltage electrolytics that you keep in a drawer by themselves, might be a few thousand microfarads.

Take any capacitor you own, charge it up to 3.46 volts, and then connect it to a night-light bulb, and don't blink . . . the bulb will give a little bleep of a flash, and that will be IT.  

The '4 farad' capacitor that answers this question is something that would have required a parallel combination of every capacitor I ever owned during the era when I had a workbench and spent any time at it.  This monstrosity would have taken days to assemble, would have occupied the whole basement, and still would likely have fallen short of whole farads.

But then, I admit to being a dinosaur. And today, when I go to FLoogle and search "supercapacitor", there they are !  Whole farads, that can stand up to a few volts, for under five bucks.  So the only thing left for me to be amazed about is the fact that this should no longer amaze me.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft