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A particle is thrown with velocity u mak...

A particle is thrown with velocity u making an angle `theta` with the vertical. It just crosses the top of two poles each of height h after 1 s and 3 s respectively. The maximum height of projectile is

A

9.8 m

B

19.6 m

C

39.2 m

D

4.9 m

Text Solution

Verified by Experts

The correct Answer is:
B
Using the relation, `s = ut + 1/2 at^(2)` , we have
`h = u cos theta t _(1) -1/2 "gt" _(1)^(2) = u cos theta t_(2) -1/2 " gt"_(2)^(2)`
or `u cos theta xx 1 -1/2 xx 9.8 xx1^(2)= u cos thetaxx 3 -1/2 xx 9.8 xx 3^2`
or ` ucos theta (3-1) = 4.9 xx (9-1) = 4.9 xx 8 `
or ` u cos theta = (4.9 xx 8)/2 = 4.9 xx 4 = 19.6 m//s`
Max. height `=(u^2 cos ^(2) theta)/(2g) =((19.6)^2)/(2xx9.8 ) = 19.6 m`
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Knowledge Check

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