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 Respected Sir/Madam; Please help me with Q.2 in the attached image....please...!!!!...

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user8235304 | Student, Grade 11 | Valedictorian

Posted June 24, 2013 at 11:34 AM via web

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 Respected Sir/Madam;

Please help me with Q.2 in the attached image....please...!!!!

Answer of Q.2 is given in the page only.

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llltkl | College Teacher | Valedictorian

Posted June 24, 2013 at 2:21 PM (Answer #1)

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For a projectile thrown with initial velocity u, and at an angle `theta` with the horizontal, the average velocity for motion between its starting point and the highest point of its trajectory, the average velocity has to be found.

We know, velocity is displacement /time. So, average velocity is the total displacement /total time.
The initial velocity, u can be resolved into two components: one horizontal, `ucostheta` , the other vertical, `usintheta` .

At the highest point, the vertical velocity is zero. Applying (v=u+at),

`0=usintheta-g*t`

`rArr t = (usintheta)/g` 

Time to reach the highest point is `(u*sin theta) /g`

Horizontal displacement = `(u*costheta*u*sintheta)/g` = `(u^2sinthetacostheta)/g`  (as horizontal velocity remains constant throughout)
Vertical displacement =`((usintheta)/g)*((usintheta)/2)`

= `(u^2sin^2theta)/(2g)`  (as vertical velocity is a constantly decreasing one, and is zero at the top, the average during rise is half its initial value).

As displacement is a vector quantity, resultant can be obtained by squaring and adding these two displacements, and finally taking the square root.

Here,

`d = sqrt (((u^2sinthetacostheta)/g)^2 + ((u^2sin^2theta)/(2g))^2)`

= `sqrt ((u^4sin^2thetacos^2theta)/g^2 + (u^4sin^4theta)/(4g^2))`

= `sqrt((u^4sin^2theta)/(4g^2)(4cos^2theta + sin^2theta))`

= `(u^2sintheta)/(2g) sqrt(cos^2theta + sin^2theta+3cos^2theta)`

= `(u^2sintheta)/(2g) sqrt(1+3cos^2theta)`

Average Velocity =` ((u^2sintheta)/(2g) sqrt(1+3cos^2theta))/((usintheta)/g) `

= `u/2 sqrt(1+3cos^2theta)`

Hence correct answer is option B).

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