A projectile is fired at an angle of 30^{circ | vedclass

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Question

(d)

$	an 30^{\circ}=\frac{u_y}{u_x}$

$	herefore \quad \frac{1}{\sqrt{3}}=\frac{80}{u_x}$

or $\quad u_x=\frac{80}{\sqrt{3}}\,m / s$

$T=\f

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Vedclass · Answer

(d)

$	an 30^{\circ}=\frac{u_y}{u_x}$

$	herefore \quad \frac{1}{\sqrt{3}}=\frac{80}{u_x}$

or $\quad u_x=\frac{80}{\sqrt{3}}\,m / s$

$T=\frac{2 u_y}{g}=\frac{2 	imes 80}{10}=16\,s$

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 $v_y=80+(-10)(4)=40\,m / s$

$	herefore \quad$ Speed $=\sqrt{\left(80 \sqrt{3)}^2+(40)^2ight.}=\frac{40}{\sqrt{13}}\,m / s$

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 $........\frac{m}{s}$

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