Nona Z.

asked • 10/10/16

Important Kinematics Physics

a-  A particle is projected vertically upwards with initial speed u m.s-1 , after T s it reaches its greatest height , H m .
Find , in terms of u and g, an expression for :
 
i - T                 ii - H
 
b- At a certain time the particle is travelling upwards through a point A,at a height of 19.6 m, and 2 s later it is at a point B, at a height of 29.4 m .
 
i- show that it is on its way down through B at that time .
 
ii- Find the value of u .

1 Expert Answer

By:

Nona Z.

Thank you very much Sir,
 
related to part B , 
what I said is the particle has travelled 19.6 m then ''Suppose we set the timer here'' he travelled 29.4 in 2 S , but the particle has already cut 19.6 m before the timer so the real distance through this time is :
29.4 - 19.6 = 9.8 m , which is the distance that the particle has travelled in 2 S ,
I then substitute the t and s and a in ;
S = ut + at2
 
as I put that a is - 9.8 m.s-2  as to reach that distance it has to be minus ' opposite  direction of gravity'
I got u = 15 m.s-1
 
I then substitute all the information that I have at ;
 
v2 = u2 + 2as 
 
 
so I got that v = 20 m.s-1 
 
As I also considered logically for the particle when its way down a = +a  , 
 
but how can I show that it is on its way down ?
I hope that my way of thinking makes physical sense .
 
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10/11/16

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