Question

A stone dropped from certain point crosses the top and bottom ends of a vertical pole of length 20m in one second the distance of top of the pole from the initial point of release will be (g=10)
Ans:11.25m

Answer 1

as it crosses pole
average speed = -20 m/s

change in speed in one second
= a(1) = -10 m/s
so
speed at top of pole = -15 m/s
so how far did it fall to get to -19.5

-15 = -10 t
t = 1.5
d= (1/2) g t^2 = 5(1.5)^2
19.361
h = (1/2)g t^2
= 5 * 2.25
= 11.25

Answer 2

d = V*t + 0.5g*t^2.
20 = V*1 + 0.5*10*1^2,
V = 15 m/s = Velocity at top of pole.

V^2 = Vo^2 + 2g*d.
15^2 = 0 + 20d,
d = 11.25 m.

Answer 3

S=vt + 1/2at²
20= v(1)+1/2(10)(1)²
20= v+5
v=15
v²=u²+2as
(15)²=(0)²+2(10)d
d=11.25