A particle is projected upwards with a velocity of $100 m / s e c$ at an angle of $60^{0}$ with the vertical. The time (in sec) when the particle will move perpendicular to its initial direction is 10x. (taking $g = 10 m / s e c^{2}$ ). The value of x is:

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Solution

$\to u = 100 s i n 60^{0}^i + 100 c o s 60^{0}^j = 50^j + 50 \sqrt{3}^i$
$\to v = \to u - g t^j = 50^j + 50 \sqrt{3}^i - 10 t^j = 50 \sqrt{3}^i + (50 - 10 t)^j$
When velocity is perpendicular to initial velocity, $\to v . \to u = 0$ , or $(50 \sqrt{3}^i + (50 - 10 t)^j) . (50^j + 50 \sqrt{3}^i) = 0$ or, $7500 + 2500 - 500 t = 0$
which gives, $t = 20 s$

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A particle is projected upwards with a velocity of 100m/sec at an angle of 600 with the vertical. The time (in sec) when the particle will move perpendicular to its initial direction is 10x. (taking g=10m/sec2). The value of x is:

→u=100sin600^i+100cos600^j=50^j+50√3^i→v=→u−gt^j=50^j+50√3^i−10t^j=50√3^i+(50−10t)^jWhen velocity is perpendicular to initial velocity, →v.→u=0, or (50√3^i+(50−10t)^j).(50^j+50√3^i)=0 or, 7500+2500−500t=0which gives, t=20s

A particle is projected upwards with a velocity of $100 m / s e c$ at an angle of $60^{0}$ with the vertical. The time (in sec) when the particle will move perpendicular to its initial direction is 10x. (taking $g = 10 m / s e c^{2}$ ). The value of x is: