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Grade 9Mechanics

A particle is thrown with a velocity u. At t=2 s the particle makes an angle 30 degree with the horizontal. At t=3 s it reaches the maximum height. Find the angle of projection and u.

Sayak

10 Years agoGrade 9

1 Answer

arun

10 Years ago

let initial velocity of projectile be ‘ $u$ ‘. And angle of projection be $\theta$

so initial velocity in horizontal dirction is $u\cos\theta$

and initial velocity in vertical direction is $u\sin\theta$

velocity of projectile at t=2 sec is $v$

firstly considering the vertical motion

at t=3 sec

from first equation of motion

$0= u\sin\theta -g(3)$

$u=\frac{3g}{\sin\theta}$ …....1

at t= 2 sec

from first equation of motion

$v\sin30=u\sin\theta-g(2)$

from equation 1

$\frac{v}{2}=3g-2g$

$v=2g$ …......2

now considering the horizontal motion

at t=2 sec

from first equation of motion

$v\cos30=u\cos\theta-(0)(2)$

from equation 2

$u=\frac{\sqrt{3}g}{\cos\theta}$

from equation 1

$\frac{\sqrt{3}g}{\cos\theta}=\frac{3g}{\sin\theta}$

$\tan \theta=\sqrt{3}$

$\theta=60$

from equation 1

$u=2\sqrt{3}g$

hence u= 2(3)^(1/2)g and angle of projection is 60

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