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Standard XII

Mathematics

Applications of Dot Product

From an eleva...

Question

From an elevated point P, a stone is projected vertically upwards. When it reaches a distance $d$ below P, its velocity is doubled. The greatest height reached by it above P is :

d / 3

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2 d

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d / 2

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Solution

The correct option is B $d / 3$
$v - u = g t$
$2 u - u = g t$
$\Rightarrow t = \frac{u}{g}$ where, t is the time to go from u (downward) to 2u (downward).
$t_{t o t a l} = \frac{2 u}{g} + \frac{u}{g} = \frac{3 u}{g}$ since $\frac{u}{g}$ is time to go up and same time to come down.
$(2 u)^{2} - u^{2} = 2 g d$
$\Rightarrow d = \frac{3 u^{2}}{2 g}$
Maximum height, $H = \frac{u^{2}}{2 g} = \frac{d}{3}$

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From an elevated point P, a stone is projected vertically upwards. When it reaches a distance d below P, its velocity is doubled. The greatest height reached by it above P is :

The correct option is B d/3v−u=gt2u−u=gt⇒t=ug where, t is the time to go from u (downward) to 2u (downward).ttotal=2ug+ug=3ug since ug is time to go up and same time to come down.(2u)2−u2=2gd⇒d=3u22gMaximum height, H=u22g=d3

From an elevated point P, a stone is projected vertically upwards. When it reaches a distance $d$ below P, its velocity is doubled. The greatest height reached by it above P is :