<span>A= a=99/85.3
B= a=-54/85.3
C=The acceleration is smaller.
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Answer:
The height of the cliff is 90.60 meters.
Explanation:
It is given that,
Initial horizontal speed of the stone, u = 10 m/s
Initial vertical speed of the stone, u' = 0 (as there is no motion in vertical direction)
The time taken by the stone from the top of the cliff to the bottom to be 4.3 s, t = 4.3 s
Let h is the height of the cliff. Using the second equation of motion in vertical direction to find it. It is given by :



h = 90.60 meters
So, the height of the cliff is 90.60 meters. Hence, this is the required solution.
With arms outstretched,
Moment of inertia is I = 5.0 kg-m².
Rotational speed is ω = (3 rev/s)*(2π rad/rev) = 6π rad/s
The torque required is
T = Iω = (5.0 kg-m²)*(6π rad/s) = 30π
Assume that the same torque drives the rotational motion at a moment of inertia of 2.0 kg-m².
If u = new rotational speed (rad/s), then
T = 2u = 30π
u = 15π rad/s
= (15π rad/s)*(1 rev/2π rad)
= 7.5 rev/s
Answer: 7.5 revolutions per second.
Pressure = Force/ Area = 3000/2 = 1500 pascal.
B is the answer, I’m really good at this subject