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Mechanics

a stone thrown upward with a speed u from the top of a tower reaches the ground with a velocity 4u. The height of the tower is .........

Swati

12 Years agoGrade

3 Answers

Arun Kumar

12 Years ago

final speed=4u=u+gt=>t=3u/g

now h=u*3u/g+0.5+g+(3u/g)^2

7.5*u^2/g

Vishal Gupta

9 Years ago

v=u+atu=+vv=-3va=-g=>-4v=-gt=>t=4v/gtime for velocity to be -v= 2v/gSo 4v/g-2v/g time is taken to cover the tower with a velocity v downwardss= -v*(2v/g) +1/2*(-g)*(2v/g)*(2v/g)=> s= -2(v^2)/g -1/2*(4(v^2)/g)=> s=-2(v^2)/g -(2(v^2)/g)=>s=-4(v^2)/gSo ht of tower is 4(v^2)/g

Abhishek

8 Years ago

u(initial)=uV(final)=-4ua=-gs=-hv^2-u^2=2as16u^2-u^2=2(-g) (-h) 15u^2=2ghh=15u^2/2g.h=7.5u^2/gThis is the height of the tower.

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