Answer:
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Answer:
It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.
V^2=U^2+2*a*x
V=0 (as it is at max height)
U=30ms^-1 (initial speed)
a=-g /-9.8ms^-2 (as it is moving against gravity)
x is the variable you want to calculate (height)
0=30^2+2*(-9.8)*x
x=-30^2/2*-9.8
x=45.92m
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Answer: Velocity terminal = 0.093m/s
Explanation:
1. We start by evaluating the gap distance between the two cylinders as h = R(sleeve) - R(cylinder)
= (0.0604/2 - 0.06/2)m
= 2×10^-4
Surface are of the cylinder in the drop, which is required in order to evaluate the shearing stress can be expressed as A(cylinder) = π.d.L
= (π×0.06×0.4)m²
= 0.075m²
Since the force of the cylinder's weight is going to balance the shearing force on the walls, we can express the next equation and derive terminal velocity from it.
Shearing stress = u×V.terminal/h = 0.86×V/0.0002
= 4300Vterminal
Therefore, Fw = shearing stress × A
30N = 4300Vterminal × 0.075
V. terminal = 30/4300 m.s
V. terminal = 0.093m/s
