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The time of flight of a projectile is 10...

The time of flight of a projectile is 10 s and range is 500m. Maximum height attained by it is [`g=10 m//s^(2)`]

A

125m

B

50m

C

100m

D

150m

Text Solution

Verified by Experts

The correct Answer is:
A
Time of flight, `T = (2u sin theta)/(g) = 10s`
`rArr u sin theta = 50 ms^(-1)`
`:. H = (u^(2)sin^(2)theta)/(2g) = ((u sin theta)^(2))/(2g) = (50 xx 50)/(2 xx 10) = 125m`
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