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3-2.Motion in Plane
hard
A particle is thrown with a speed of $12\, m/s$ at an angle $60^o$ with the horizontal. The time interval between the moments when its speed is $10\, m/s$ is $(g = 10\, m/s^2)$.........$s$
A$1$
B$1.2$
C$1.4$
D$1.6$
Solution
$v_{H}=u \cos \theta=6$
$\mathrm{v}_{\mathrm{v}}=\sqrt{\mathrm{v}^{2}-\mathrm{u}^{2} \cos ^{2} \theta}=8$
$t_{1}=\frac{u \sin \theta-8}{10}$
$\mathrm{t}_{2}=\frac{\mathrm{u} \sin \theta+8}{10}$
$\mathrm{t}_{2}-\mathrm{t}_{1}=\frac{8 \times 2}{10}=1.6 \mathrm{s}$
$\mathrm{v}_{\mathrm{v}}=\sqrt{\mathrm{v}^{2}-\mathrm{u}^{2} \cos ^{2} \theta}=8$
$t_{1}=\frac{u \sin \theta-8}{10}$
$\mathrm{t}_{2}=\frac{\mathrm{u} \sin \theta+8}{10}$
$\mathrm{t}_{2}-\mathrm{t}_{1}=\frac{8 \times 2}{10}=1.6 \mathrm{s}$
Standard 11
Physics
