KVPY SA Exam 2011 Question Paper

Question : 18 of 80

Marks: +1, -0

A juggler tosses a ball up in the air with initial speed u. At the instant it reaches its maximum height H, he tosses up a second ball with the same initial speed. The two balls will collide at a height

$H/4$
$H/2$
${3H}/4$
$√{3/4}H$

Validate

Solution:

H-h=1/2gt^2

h=ut-1/2gt^2

H = ut

…..(1)

u^2/{2g}=H

…..(2)

t=u/{2g}

(from (1) and (2))

∴h=u×u/{2g}-1/2g×u^2/{4g^2}

={3u^2}/{8g}={3H}/4

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