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marysya [2.9K]
3 years ago
5

a rescue line is to be thrown horizontally from the bridge of a cruise ship at 40.0 m above sea level to a life boat 50.0 m away

with what horizontal speed should the rescue line be thrown?
Physics
1 answer:
Gemiola [76]3 years ago
8 0
There is a 40m drop from the bridge, so we can work out the time of the vertical component first (how long it will take to reach the water) 

do this by using a SUVAT equation s=ut+0.5at^2
this rearranges to t= sqrt(2s/a)
- note that u has been removed since there is no initial velocity.
 
sqrt((2*40)/9.81)= 2.86 seconds 

this means that the horizontal component of 50m has so be covered in 2.86s 
and speed = distance/ time 

50m/2.86s=17.48ms^-1

the line has to be thrown at 17.48ms^-1 or more to reach the life boat. 
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This question is based on the fundamental assumption of  vector direction.

A vector is  a physical quantity which has  magnitude as well direction  for its complete specification.

The magnitude of a physical quantity is simply a  numerical number .Hence it can not be negative.

A negative vector is a vector which comes into existence when it is opposite to our assumed direction with respect to any other vector.  For instance, the vector is taken positive if it is along + X axis and negative if it is along - X axis.

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The second option tells that the magnitude of the vector is less than 1. Magnitude can not be negative. So this is also wrong.

Third one tells that a vector is negative if its displacement is along north. It does not give any detail information about the negativity of a vector.

In a general sense we assume that vertically downward motion  is negative and vertically upward is positive. In case of a falling object the motion is  vertically downward. So the velocity of that object is negative .

So last   option is  partially  correct  as  the vector can be negative depending on our choice of co-ordinate system.





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