A particle is thrown vertically up with velocity 10 ms from a point P which is 10 m above the ground. Time after which particle reaches the ground. (g=10 ms2)

h=−vt+12gt2⇒10=−10t+12×10t2⇒5t2−10t−10=0 ⇒t2−2t−2=0 t=(3+1)sec

A particle is thrown vertically up with velocity 10 ms from a point P which is 10 m above the ground. Time after which particle reaches the ground. (g=10 ms2)

h=−vt+12gt2⇒10=−10t+12×10t2⇒5t2−10t−10=0 ⇒t2−2t−2=0 t=(3+1)sec

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A particle is thrown vertically up with velocity $10 \frac{m}{s}$ from a point P which is 10 m above the ground. Time after which particle reaches the ground. $(g = 10 \frac{m}{s^{2}})$

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$(\sqrt{3} - 1) S$

$(\sqrt{3} + 1) S$

2 S

3 S

answer is B.

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Detailed Solution

$h = - v t + \frac{1}{2} g t^{2} \Rightarrow 10 = - 10 t + \frac{1}{2} \times 10 t^{2} \Rightarrow 5 t^{2} - 10 t - 10 = 0 \Rightarrow t^{2} - 2 t - 2 = 0 t = (\sqrt{3} + 1) s e c$

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A particle is thrown vertically up with velocity 10 ms from a point P which is 10 m above the ground. Time after which particle reaches the ground. (g=10 ms2)

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A particle is thrown vertically up with velocity $10 \frac{m}{s}$ from a point P which is 10 m above the ground. Time after which particle reaches the ground. $(g = 10 \frac{m}{s^{2}})$