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In the projectile motion shown is figure...

In the projectile motion shown is figure, given `t_(AB) = 2s then (g = 10 ms^(-2))`

A

particle is at point B at 3 s

B

maximum height of projectile is 20 m

C

initial vertical component of velocity is `20 ms^(-1)`

D

horizontal component of velocity is `20 ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
Horizontal component of velocity remains unchanged
`X_(OA) = 20 m = X_(AB)/2`
`:. t_(OA) = t_(AB)/2 = 1s`
For AB projectile
`T = 2 s = (2u_y)/g`
`:. u_y = 10m//s`
`H = (u_(y)^2)/(2g) = (10)^2/(2 xx10) = 5m`
`:.` Maximum height of total projectile,
`= 15 + 5 = 20 m`
`t_(OB) = t_(OA) + t_(AB) = 1+2= 3s` .
For complete projectile
`T = 2(t_OA) + t_(AB)`
` = 4 s = (2u_y)/g`
`:. u_y = 20 m//s`
`u_x = (AB)/(t_(AB)) = 40/2 = 20 m//s`.
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