PHYSICS question about speed.? PLEASE SHOW WORK!!?

Question

A falling stone takes 0.30 s to travel past a window 2.2 m tall. From what height above the top of the window did the stone fall?

PLEASE show all calculations. thanks a bunch.

Answer 1

t = 0.3 sec
d = 2.2 m
so, average speed = 2.2/0.3 = 7.33 m/sec
at t=0.3, the change in velocity= gt = 9.8(0.3) = 2.94 m/sec.
Therefore, the speed st the top of the window = 7.33 - 2.94/2
or 5.86 m/sec.
The stone had traveled vf = vi + gt
5.86 = 0 + 9.8t
t = 0.598 sec - the time the stone is already falling before the window.
D = vt/2 = 5.86(0.598)/2 = 1.752 m above the window.

Answer 2

distance at ascelerated motion is
s = a t^2 / 2 (a - asceleration, here=9.8m/s^2; t=time)
let the stone be falling for t0 sec before reaching window. It travelled s0 = a t0^2 / 2 meters. 0.30s later we have
s0 + 2.2 = a (t0+0.30)^2 / 2. from these equations follows
a t0^2 / 2 + 2.2 = a (t0+0.30)^2 / 2
a t0^2 / 2 + 2.2 = a t0^2 / 2 + a t0 0.30 + a 0.30^2 / 2
t0 0.30 + 0.30^2 / 2 = 2.2 /a
t0 = 2.2 / (a 0.30) - 0.30 / 2
t0 = 2.2 / (9.8 0.30) - 0.30 / 2 ~ 0,5983 s
and s0 = 9.8 t0^2 / 2 = 1,7540 m

sorry for possible spell errors

Answer 3

h = (1/2)at^2
h + 2.2 = (1/2)a(t + 0.3)^2
2.2 = (1/2)a[(t + 0.3)^2 - t^2]
t^2 + 0.6t + 0.09 - t^2 = 4.4/9.80665
6t + 0.9 = 44/9.80665
t = (44/9.80665 - 0.9)/6
h = (1/2)(9.80665)[(44/9.80665 - 0.9)/6]^2
h ≈ 63.080 m