Physics: How to solve this problem are problems like it?

Since it is dropped near earth's surface (assumed), we know A=9.8m/s^2.

We also know that Vi=0 since we assume it is not moving before being dropped.

We also know T=15s because it was given.

Now we look at our equations. Which one gives us those 3 variables so that we can find something we want?

That's right, it's X=ViT+1/2AT^2, which gives us distance, or depth in this case.

Once you've found X, you want to plug it into Vf^2=Vi^2+2AX and solve for Vf. That will give you the final velocity at time of impact.

Yes, all this looks so confusing, doesn't it!

Get simple answers first, then check that with "complicated looking" equations.

Velocity at bottom = (gt), = 9.8 x 15, = 147m/sec. (use Vf = Vi + AT), where Vi = 0, and A = g (9.8).

Depth = 1/2 (t^2 x g) = 1,102.5 metres. (Use X = ViT + 1/2 AT^2). Vi does not exist, so

X = 1/2 A(T^2), where again A is g. (9.8).

I'm sure teachers are trying to bamboozle students by making things look difficult!

i'd say use x=vit+1/2at^2 to find distance b/c you know accel 9.8

then use vf^2=vi^2+2ax to get your final velocity

I would try to reason it out - formulas work but U shouldn't use them w/o understanding.

Every object that falls freely (called Free-Fall) here on Earth

gathers speed at a rate of 9.8 meters/second (m/s) - every second it falls.

So if an object falls for 15 s, it obtains a speed = 15 x 9.8 = 147 m/s ANS

and

the distance an object Free-Falls is always the average speed x time it falls.

In this case,

the average speed is one-half of the final speed which = 1/2(147) = 73.5 m/s

therefore

distance = 73.5 x 15 = 1103 m ANS