Physics help pls.....give me correct ans. & procedure & i'll give u whole 10 pts...?

Question

1. The empire state building is 1472 ft. high.
a. how long will it take an object drop from the top of the
building to reach the ground??
b. what would an objects final velocity be??

2. A body falls freely from the rest. FIND:
a. the acceleration
b. the distance that falls 3s
c. it's speed after falling 70m
d. time required to reach the speed of 25m/s
e. time taken to fall 300m

Answer 1

Formula:
A.
S = V1(t) + 0.5gt^2
Initial velocity = 0
1472 ft = 0 + 0.5(32 ft/s^2)(t)^2
t = sqrt((1472)/((0.5)(32))
t = 9.59 seconds


B.
V2 = V1 +gt
V1 = 0
V2 = 32ft/s^2X9.59 s
V2 = 306.88 ft/s

2.a ) If S.I: 9.8m/s^2, If English : 32 ft/s^2
b) s = 0.5gt^2
S.I :
s = 0.5(9.8)(3)^2
s = 44.1 m
English
s = 0.5(32)(3)^2
s= 144 ft
c) s = 0.5gt^2
70 = 0.5(9.8)(t)^2
t = sqrt((70)/(0.5X9.8))
t = 3.77 seconds
d)V2 = gt
25 = (9.8)(t)
t = 25/9.8 = 2.55 seconds
e) s= 0.5gt^2
300 = 0.5(9.8)(t)^2
t = sqrt((300)/(0.5X9.8))
t = 7.82 seconds

Answer 2

1.)
a.) t =√(2s/g) = √((2)(1472)/(32)) = 9.592 s
b.) v = at = gt = (32)(9.592) = 306.93 ft/s
2.)
a.) 9.8 m/s
b.) s = .5gt^2 = (.5)(9.8)(3^2) = 44.1 m
c.) t = √(2s/g) = √(2(70)/9.8) = 3.7796 s
v = at = gt = 9.8(3.7796) = 37.04 m/s
d.) t = v/a = 25/9.8 = 2.551s
e.) t = √(2s/g) = √((2)300/9.8) = 7.8246s

Answer 3

supermontero2000 beat me to it.

Agreeing with all his answers.