A particle is thrown with velocity u making anangle with the vertical.

1.	A particle is thrown with velocity u making anangle with the vertical. It just crosses the topof two poles each of height h after 1 s and 3srespectively. The maximum height of projectile is
Answer» ANSWER: EXPLANATION: tH=v0yg and H=v20y2g=(gtH)22g=12gt2H Here t H=(1+3)2=2s Hence H=12(9.8)(2)square =19.6m

A particle is thrown with velocity u making anangle with the vertical. It just crosses the topof two poles each of height h after 1 s and 3srespectively. The maximum height of projectile is

Answer»

ANSWER:

EXPLANATION:

tH=v0yg and

H=v20y2g=(gtH)22g=12gt2H

Here t H=(1+3)2=2s

Hence H=12(9.8)(2)square =19.6m

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A particle is thrown with velocity u making anangle with the vertical. It just crosses the topof two poles each of height h after 1 s and 3srespectively. The maximum height of projectile is

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