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
Equation 1 :
m1 : m1a =T
Equation 2:m2 : m2a= F - T
Adding 1 and 2a=0.073
Placing in equation 1
we get T = 218.18N
Nuclear fusion in the core tries to blow the star apart. Gravity holds it together. Whoever designed that system really knew what he was doing. I'm kinda grateful to him.
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