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Prove that when a shot is projected from a gun at any angle of elevation, the  shot as seen from the point of projection will appear to descend past vertical  target with uniform velocity. 18. A ball is projected with a velocity u at an elevation a from point distance from  a smooth vertical wall, it returns to the point of projection. Prove that 2 sin 2 gd e u gd    where e= coefficient of restitution. Hence find the maximum d  for which the ball can return to the point of projection.
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Asked by dhandas060104 | 25 Oct, 2020, 11:53: AM
answered-by-expert Expert Answer
I do not understand Qn. No.17 , how the ball falls under gravitational field appears as falling with constant speed.
 
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Qn. No. 18
 
if a ball is projected with initial velocity u with projection angle α , time taken t to travel horizontal distance d is given as
 
t = d / ( u cosα ) ...............(1)
 
Vertical distance travelled in above time t is determined from the equation ,
 
h = [ ( u sinα ) t ] - [ (1/2) g t2 ]  ................(2)
 
By substituting t from eqn.(1) , we rewrite eqn.(2) as
 
begin mathsize 14px style h space equals space d space tan alpha space minus space fraction numerator g space d squared over denominator 2 space u squared cos squared alpha end fraction end style   ................... (3)
vertical component velocity uv after time t as mentioned in eqn.(1) is , 
 
uv = ( u sinα ) - ( g t ) = ( u sinα ) - [ ( gd ) / ( u cosα ) ]
 
uv = [ ( u2 sin2α ) - ( 2 g d ) ] / ( 2 u cosα ) ..............................(4)
 
After hitting the opposite wall the ball is reflected  so that its magnitude of horizontal and vertical components are
multiplied by e which is coefficient of restitution .  
 
It is given that ball returns back to same projection point . 
 
Time taken t' to travel horizontal distance d during reverse motion 
 
t' = d / ( e u cosα ) .....................(5)
 
Vertical distance travelled during this time given in eqn.(5) 
 
h = ( -e uv ) t' + (1/2) g t' 2 ..................(6)
 
By substituting uv from eqn.(4) and t'  from eqn.(5), we rewrite eqn.(6) after simplification as
 
begin mathsize 14px style h space equals fraction numerator g d squared over denominator u squared cos squared alpha end fraction space minus space d space tan alpha space plus space 1 half fraction numerator g d squared over denominator e squared u squared cos squared alpha end fraction end style  .....................(7)
Since the vertical distances travelled in forward motion and backward motion are same ,
 
we equate eqn.(3) and (7) and after simplification , we get
 
begin mathsize 14px style 2 space tan alpha space equals 3 over 2 fraction numerator g d over denominator u squared cos squared alpha end fraction space plus space 1 half fraction numerator g d over denominator e squared u squared cos squared alpha end fraction end style
 
begin mathsize 14px style 2 space tan alpha space equals fraction numerator g d over denominator 2 u squared cos squared alpha end fraction open parentheses 3 plus 1 over e squared close parentheses space end style
 
begin mathsize 14px style fraction numerator 4 space u squared space cos squared alpha space tan alpha over denominator g d end fraction space equals space 3 space plus space 1 over e squared end style
 
begin mathsize 14px style fraction numerator 2 space u squared sin 2 alpha space minus space 3 g d over denominator g d end fraction space equals space 1 over e squared end style
begin mathsize 14px style e squared space equals space fraction numerator g d over denominator 2 u squared sin 2 alpha space minus space 3 g d end fraction end style
Answered by Thiyagarajan K | 25 Oct, 2020, 09:14: PM

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