Engineering Mechanics - PKRB: Impulse and Momentum - Discussion

Discussion Forum : PKRB: Impulse and Momentum - General Questions (Q.No. 1)
1.

The uniform rod AB has a weight of 3 lb and is released from rest without rotating from the position shown. As it falls, the end A strikes a hook S, which provides a permanent connection. Determine the speed at which the other end B strikes the wall at C.

vB4 = 14.74 ft/s
vB4 = 20.3 ft/s
vB4 = 22.0 ft/s
vB4 = 12.69 ft/s
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
4 comments Page 1 of 1.

DKS said:   7 years ago
A is correct by considering 3.5ft.

Adarsh said:   9 years ago
As per my calculation, it's 14.74.

So the answer is option A.

Nishant said:   9 years ago
Can anybody solve this?

Supratik said:   1 decade ago
How this possible ? can you please explain me.

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