A projectile is fired at an angle of 30^{circ} to horizontal such that vertical component of its ini

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3-2.Motion in Plane

hard

A projectile is fired at an angle of $30^{\circ}$ to the horizontal such that the vertical component of its initial velocity is $80\,m / s$. Its time of flight is $T$. Its velocity at $t=\frac{T}{4}$ has a magnitude of nearly $........\frac{m}{s}$

A

$200$

B

$300$

C

$100$

D

None of these

Solution

(d)

$\tan 30^{\circ}=\frac{u_y}{u_x}$

$\therefore \quad \frac{1}{\sqrt{3}}=\frac{80}{u_x}$

or $\quad u_x=\frac{80}{\sqrt{3}}\,m / s$

$T=\frac{2 u_y}{g}=\frac{2 \times 80}{10}=16\,s$

At $\quad t=\frac{T}{4}=4\,s$

$v_x=u_x=\frac{80}{\sqrt{3}}\,m / s$

and $v_y=u_y+a_y t$

Or $v_y=80+(-10)(4)=40\,m / s$

$\therefore \quad$ Speed $=\sqrt{\left(80 \sqrt{3)}^2+(40)^2\right.}=\frac{40}{\sqrt{13}}\,m / s$

Standard 11

Physics

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