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Blazing goIITian

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2 Jul 2009 19:19:20 IST

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A q on projectile motion

Kinematics , NewtonsLawsCirMotion , Work Power Energy collisions , Rotation , Gravitation , Simple Harmonic Motion , Fluid Mechanics , Materials Wave Sound , Super Position , units dimensions , com momentum , shm , wave sound

A body is thrown horizontally,from the top of a tower and it strikes the ground after t seconds at an angle 45^{0 with the horizontal.Find the height of the tower and speed of projection}

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Comments (8)

Abhinav Jain

Blazing goIITian

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2 Jul 2009 19:22:36 IST

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i got the height H=45m but plz explain for velocity

Abhinav Jain

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2 Jul 2009 19:23:13 IST

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i also got the y component of v as -30 but still not able to get the x component

sarang

Blazing goIITian

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2 Jul 2009 19:23:20 IST

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h=1/2gt^2

v=gt

Abhinav Jain

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2 Jul 2009 19:36:50 IST

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how v=gt?

Shrey Choudhary

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2 Jul 2009 19:54:35 IST

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after t secs,H=1/2gt^2 you'll get H from here.Since Uy initial =0 and since angle of projection becomes 45 deg after time t Vy / Vx = tan 45 which implies Vy=Vx which further implies ,after time t, Vy=Vx(at time t)=Ux(Initial)...coz there is no acceleration in X direction.Then Vy=Uy + gt and Uy=0.....Vy=gt=Ux=Speed of projection....Rate if useful

asdfg

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2 Jul 2009 22:54:00 IST

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Abhinav can you please explain how you got h=45m.

Abhinav Jain

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3 Jul 2009 09:29:25 IST

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ya srry i forgt to mention in the q at t=3s,so substituting t=3 in h=5t2 we get h=45m

~Áß???????

Blazing goIITian

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3 Jul 2009 10:29:18 IST

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yup..

now as t = 3sec u mentioned..!

after 3 sec..

the distance travelled by particle vertically = = 45 m.

but at pt it hit the ground ..

the angle made = 45m so now considering figure as angle is 45 deg ..

so tan45 = 1 hence horizontakl distance is also 45 m

and time 3 sec.

so v = 45/3 = 15 m/sec.!

(IT BECOME TOO EASY AFTER THAT INFO)

BEST OF LUCK..

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