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Q.Two particles A and B are projected from the top of a tower of height 80 m with same...

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user8235304 | Student, Grade 11 | Valedictorian

Posted July 20, 2013 at 12:45 PM via web

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Q.Two particles A and B are projected from the top of a tower of height 80 m with same speed 20 m/s. A is thrown at an angle 30° above the horizontal line and B is thrown at an angle 30° below the horizontal line.They will hit the ground:-

a) in same time

b) with same speed

c) with same horizontal range

d) time taken by A to hit the ground is less than that by B.

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llltkl | College Teacher | Valedictorian

Posted July 20, 2013 at 7:29 PM (Answer #2)

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For particle A,

Time to reach the peak, `t=20*sin30/g=10/10=1 s` (assuming the value of g to be `10 m/s^2` ).

If H be the maximum height attained by the particle, then considering vertical component,

`0= (20*sin30)^2-2*10*H`

`rArr H=5 m`

Total height of the particle from ground level, at the peak of its flight =80+5=85 m

Time to reach the ground from its peak =`sqrt(2*85/10)=4.12 s`

Total time of its flight`=4.12+1=5.12 s` .

Horizontal range=`20*cos30*5.12=88.68 m` .

Final velocity, at the moment of hitting the ground, `v=sqrt(0+2*10*85)=41.23 m/s`

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For particle B,

Considering vertical downward motion,

`80=20*sin30*t+1/2*10*t^2`

`rArr t^2+2t-16=0`

`rArr t=3.12 s`

Time to reach the ground, i.e. its total time of flight=`3.12 s`

Horizontal range=`20*cos30*3.12=54.04 m`

Final velocity, at the moment of hitting the ground, `v=sqrt((20*sin30)^2+2*10*80)`

`sqrt(1700)=41.23 m/s`

Both the particles hit the ground with same velocity.

Therefore, option B) is the only correct answer.

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