S = ut + at²/2, u=0
S = (32)(5)²= 32*25 = 800
Therefore, after 5 seconds, the stone is (1050-800)=250 feet above the ground.
When S=1050, it hits the ground.
Therefore, 1050=(32)*(t²)
t² = 1050/32 = 32.8125
t = 5.7822 Seconds (approximately)

I think ganesh forgot about the '/2' bit in his formula, so got the distance travelled in A double what it should have been, and that error carried forward into B.

Actually, Ganesh forgot about the initial velocity that the stone was thrown.

Mathyperson,

     Air resistance would not be small in this case unless the stone was much more dense than normal rock because the v^2 term in the air resistance formula would be very large nearing impact versus the force of gravitation.
That is nearly 12000(cross sectional area of object) newtons of resistance for a perfectly smooth round surface with zero humidity!  I'm thinking this object would have reached terminal velocity well short of impact.
     That would change both answers significantly.

1 inch stone reaches terminal velocity at about 23.93m/s (m=18.9g)
4 inch stone reaches terminal velocity at about 47.86m/s (m=1.2kg)

Our example would suggest that this stone would be traveling at 79.2m/s
A 10.9 inch rock with a mass of 24.74kg would be necessary to achieve this velocity!
I would not recommend throwing a 55 pound anything from that height.

I thought that I would drop a line to show how dramatic air resistance is in everyday situations.

My contribution for anyone that cares:

     terminal velocity = {[8gr(density of object)] / 3(density of medium)}^(1/2)

where: g = -9.8m/s
           r =  radius of object
           object must be smooth and round for this formula

     The above stone calculations were made with these densities.

     density of object = 2200kg/m^3 (based on the average density of the earth's crust)
     density of air = 1.275kg/m^3 (sea level)