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A projectile is thrown in the upward direction making an angle of $60^o $ with the horizontal direction with a velocity of $147\ ms^{-1}$ . Then the time after which its inclination with the horizontal is $45^o $ , is ......... $\sec$
$15$
$10.98$
$5.49$
$2.745$
Solution
At the two points of the trajectory during projection. the horizontal component of the velocity is the same.
$\Rightarrow 147 \times \frac{1}{2}=v \times \frac{1}{\sqrt{2}} \Rightarrow v=\frac{147}{\sqrt{2}} \mathrm{m} / \mathrm{s}$
Vertical component of
$\mathrm{u}=\mathrm{usin} 60^{\circ}=\frac{147 \sqrt{3}}{2} \mathrm{m}$
Vertical component of
$v=v \sin 45^{\circ}=\frac{147}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{147}{2} \mathrm{m}$
but $v_{y}=u_{y}+a_{y} \Rightarrow \frac{147}{2}=\frac{147 \sqrt{3}}{2}-9.8 t$
$\Rightarrow 9.8 \mathrm{t}=\frac{147}{2}(\sqrt{3}-1) \Rightarrow \mathrm{t}=5.49 \mathrm{s}$
