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3-2.Motion in Plane
hard
A projectile is thrown with a velocity of $10\,m / s$ at an angle of $60^{\circ}$ with horizontal. The interval between the moments when speed is $\sqrt{5 g}\,m / s$ is $..........\,s$ $\left(g=10\,m / s ^2\right)$.
A$1$
B$3$
C$2$
D$4$
Solution
(a)
$v^2=v_y^2+v_x^2$ or $5 g=\left(u_y-g t\right)^2+u_x^2$
$\text { or } 50 =(5 \sqrt{3}-10 t)^2+(5)^2$
$\therefore (5 \sqrt{3}-10 t) = \pm 5$
$t_1=\frac{5 \sqrt{3}+15}{10}$ and $t_2=\frac{5 \sqrt{3}-5}{10}$
$\therefore \quad t_1-t_2=1\,s$
$v^2=v_y^2+v_x^2$ or $5 g=\left(u_y-g t\right)^2+u_x^2$
$\text { or } 50 =(5 \sqrt{3}-10 t)^2+(5)^2$
$\therefore (5 \sqrt{3}-10 t) = \pm 5$
$t_1=\frac{5 \sqrt{3}+15}{10}$ and $t_2=\frac{5 \sqrt{3}-5}{10}$
$\therefore \quad t_1-t_2=1\,s$
Standard 11
Physics
