Answer:
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Answer:
so initial speed of the rock is 30.32 m/s
correct answer is b. 30.3 m/s
Explanation:
given data
h = 15.0m
v = 25m/s
weight of the rock m = 3.00N
solution
we use here work-energy theorem that is express as here
work = change in the kinetic energy ..............................1
so it can be written as
work = force × distance ...................2
and
KE is express as
K.E = 0.5 × m × v²
and it can be written as
F × d = 0.5 × m × (vf)² - (vi)² ......................3
here
m is mass and vi and vf is initial and final velocity
F = mg = m (-9.8) , d = 15 m and v{f} = 25 m/s
so put value in equation 3 we get
m (-9.8) × 15 = 0.5 × m × (25)² - (vi)²
solve it we get
(vi)² = 919
vi = 30.32 m/s
so initial speed of the rock is 30.32 m/s
Explanation:
It is given that,
Spring constant of the spring, k = 15 N/m
Amplitude of the oscillation, A = 7.5 cm = 0.075 m
Number of oscillations, N = 31
Time, t = 15 s
(a) Let m is the mass of the ball. The frequency of oscillation of the spring is given by :
Total number of oscillation per unit time is called frequency of oscillation. Here, 
m = 0.0895 kg
or
m = 89 g
(b) The maximum speed of the ball that is given by :
Hence, this is the required solution.
