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At a height of 15 m from ground velocity...

At a height of 15 m from ground velocity of a projectile is `v = (10hati+ 10hatj)`. Here `hatj` is vertically upwards and `hati` is along horizontal direction then (g = 10 `ms^(-1)`)

A

particle was projected at an angle of `45^@` with horizontal.

B

time of flight of projectile is 4 s

C

horizontal range of projectile is 100 m

D

maximum height of projectile from ground is 20 m.

Text Solution

Verified by Experts

The correct Answer is:
B, D
`u_x = v_x = 10 m//s`
`u_y = (sqrt(v_(y)^2 + 2gh))`
`= (sqrt(10^2 + (2)(10)(15)))`
`= 20m//s`
Angle of projection,
`theta = tan^(-1) (u_y/u_x) = tan^(-1) (2)`
`T = (2u_y)/g = ((2)(20))/10 = 4 s `
`R = u_xT = (10)(4) = 40m `
`H = (u_(y)^2)/(2g) = (20^2)/(2xx10) = 20 m` .
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