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At a height of 45 m from ground velocity...

At a height of 45 m from ground velocity of a projectile is,
`v = (30hati + 40hatj) m//s`
Find initial velocity, time of flight, maximum height and horizontal range of this projectile. Here `hati` and` hatj` are the unit vectors in horizontal and vertical directions.

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The correct Answer is:
A, B, C, D
Given, `v_x = 30m//s and v_y = 40 m//s `
Horizontal component of velocity remains unchanged.
`:. u_x = v_x = 30 m//s`
Vertical component of velocity is more at lesser heights.
Therefore,
`u_ygtv_y`
or `u_y = (sqrt((v_(y)^2)+2 gh))`
` = (sqrt((40)^2 + (2)(10)(45)))`
`= 50 m//s `
Initial velocity of projectile,
`u = (sqrt((u_(x)^2) +(u_(y)^2)))`
`=(sqrt((30)^2 + (50)^2))`
`=10(sqrt34) m//s `
`tan theta = u_y/u_x = 50/30 = 5/3`
` :. theta = tan^-1 (5/3)`
Time of flight, `T = ((2u sin theta)/(g)) = ((2u_y)/g)`
`= ((2xx50)/10)`
= 10s.
Maximum height, `H=(u^(2)sin^(2)theta)/(2g)=(u_(y)^(2))/(2g)`
`=((5)^(2))/(2xx10)=125m`
Horozontal range `R=u_(x)T`
`=30xx10`
`=300 m`
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