Answer: (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
