The zero gravity research facility at the nasa glenn research center includes a
145
m
drop tower. this is an evacuated vertical tower through which, among other possibilities, a
1
m
diameter sphere containing an experimental package can be dropped.

(a) how long is the sphere in free fall?

(b) what is its speed just as it reaches a catching device at the bottom of the tower?

(c) when caught, the sphere experiences an average deceleration of
29.0
g
as its speed is reduced to zero. through what distance does it travel during the deceleration?

(a) t = 5.44 sec

(b) vf = 53.31 m/s

(c) s = 5.0m

Explanation: from the question, given data

the Height of the tower, h = 145m

from question

(a)

the initial velocity, v₁ = 0 m/s

s = v₁t + 1/2 gt²

-145 m = 0(t) + 1/2 (-9.8t²)

  t² = 145/4.9

   t² = 29.59

    t = 5.44 sec

(b)

the speed of the sphere at the bottom of the tower is

vf² = vi² +2as

vf² = 0 + 2(-9.8 × -145)

vf² = 2842

vf = 53.31 m/s

(c)

when caught, the sphere experiences a deceleration of;

   a = -29.0g

the time it would take to decelerate becomes;

vf = vi + at

0 = (53.31) + (-29 ×9.8)t

where t = 53.31 / 284.2

t = 0.1876 sec

∴ the distance travelled during the deceleration becomes;

vf² = vi² + 2as

s = (vf² - vi²) / 2a

s = (0 - 53.31²) / 2×-29×9.8

s = -2841.9561 / -568.4

s = 4.99 ≈ 5.0m

i hope this helps, cheers

true is the correct answer

hope i