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Grade 12Mechanics

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Profile image of rajan jha
15 Years agoGrade 12
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1 Answer

Profile image of vikas askiitian expert
ApprovedApproved Tutor Answer15 Years ago

let its velocity  was v when it was in circular motion...after the string breaks , stone will move parabolically...

H = 2.7 (above the ground)

Ux= v       (velocity in circular path)

ay = -g

for horizontal motion threre is no accleration so ax =0

for verticle motion ,

S = gt2/2

2.7 = gt2/2

t = 0.74sec

now for horizontal motion

speed = distance/time

 Ux = 10/(0.74)                                   (distance 10m (given))

Ux = 13.51m/s

with this velocity particle was circulating .....

ac(centripital accleration) = v2/r = (13.51)2/1.5 = 122 m/s2