Grade 12th passMechanics

A body is released from a height of 300m from the ground.Exactly at the same time another body is projected from the ground in verticaly upward direction with a velocity of 159m/s along the same line.Find when and where they will meet?

V G Munj

8 Years agoGrade 12th pass

1 Answer

Gaurav Gupta

8 Years ago

we can use the relative velocity approach here. let the body at a height of 300m is body A and body at ground is body B.

lets find velocity of A with respect to B.

V_AB = V_A-V_B = 159 - 0 = 159m/s

similarly acceleration of A with respect to B

a_AB = a_A-a_B = -9.81-(-9.81) = 0

so now as B is at rest in this frame, A has to travel a distance of 300m to meet B, with “0” acceleration

so time taken to meet B will be

$T=\frac{d}{V} = \frac{300}{159} = 1.88sec$

so they will meet after 1.88sec

now coming back in ground frame distance travelled by block B in 1.88 sec

$-S = uT-\frac{1}{2}gT^{2}$ downwards is taken as negative

u =0

solving this gives

S = 17.5 m

so they will meet when B travells 17.5m or A travels 282.5m

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A body is released from a height of 300m from the ground.Exactly at the same time another body is projected from the ground in verticaly upward direction with a velocity of 159m/s along the same line.Find when and where they will meet?

V G Munj

1 Answer

Gaurav Gupta

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