Home » Ask & Discuss » Physics
« Back to Discussion

Mechanics

Hot goIITian

Joined:	16 Nov 2008
Post:	188

21 Nov 2008 14:03:53 IST

0 People liked this

625

plzz solve it. it's vert urgent

Engineering Entrance , Medical Entrance , AIPMT , JEE Main , AIIMS , JEE Advanced , Physics , Mechanics

plzz solve it. it's vert urgent

Comments (6)

Karthik M

Blazing goIITian

Joined: 1 May 2007

Posts: 2830

21 Nov 2008 14:04:32 IST

Like 0 people liked this

Joshi Bhavna

Hot goIITian

Joined: 16 Nov 2008

Posts: 188

21 Nov 2008 14:10:10 IST

Like 0 people liked this

from the base of lamp post, a particle is projected with velocity 10m/s at the angle 60 degree with horizotal. the height of lamp post is 4 times the maximum hight attained by particle. when the particle is at highest point the motion of it's shadow is observed on x-axis. then find acceleration of shadow, velocity of shadow , velocity of shadow with respect to particle and time at which particle and shadow coincide

Joshi Bhavna

Hot goIITian

Joined: 16 Nov 2008

Posts: 188

21 Nov 2008 14:14:34 IST

Like 0 people liked this

challenge for goiit'ans

Joshi Bhavna

Hot goIITian

Joined: 16 Nov 2008

Posts: 188

21 Nov 2008 14:54:00 IST

Like 0 people liked this

plzzzzzzzzzzzzz help

Harpreet aka MAX

Hot goIITian

Joined: 20 Oct 2008

Posts: 194

11 Dec 2008 20:45:11 IST

Like 0 people liked this

See the acceleration of the shadow will be Zero as we know the shadow moves in forwqrd direction (along x axis)and a projectile 's motion in horizontal direction is not accelerated and the velocity in X axis thus remains constant

Correct me if i am wrong

MAK

Hot goIITian

Joined: 19 Nov 2008

Posts: 134

11 Dec 2008 20:53:59 IST

Like 0 people liked this

dis was d solution dat i posted when u asked dis question earlier...

using equations of projectile motion for range, height and time of flight...

we get... height of lamp = 15m... distance travelled by d shadow by d tym particle reaches d max. ht. position = 10/sqrt(3) ... and d time of ascent = (sqrt3) / 2...

now using d equations of motion... we get

v = 20/3 + [(sqrt3)/4].a and u = 20/3 - [sqrt(3)/4].a ...

vel of shadow = 20/3 m/s... vel. of shadow wrt particle = 20/3 - 10(cos60) = 5/3 m/s ... and both d shadow n d particle coincide when d particle touches d ground... hence d tym taken = time of flight = sqrt(3) sec...

i think d acceleration of d shadow will be zero since it depends only on d horizantal motion of d particle and d particle has constant vel along horizantal direction... hence acc must be zero...!!!!! yet... i'm not sure of dat... but d given 3 ans are correct...

rate if useful...

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team