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A projectile is fired with a speed u at ...

A projectile is fired with a speed u at an angle `theta` above the horizontal field. The coefficient of restitution between the projectile and field is e. Find the position from the starting point when the projectile will land at its second collision

A

`(e^(2)u^(2)sin 2theta)/g`

B

`((1-e^(2))u^(2)sin 2theta)/g`

C

`((1-e)u^(2)sinthetacostheta)/g`

D

`((1+e)u^(2)sin 2theta)/g`

Text Solution

Verified by Experts

The correct Answer is:
D
Vertical velocity after collision =`"e u" sintheta`
`"So " x=R+R'=("u"^(2)sin2theta)/g+(2ucostheta("eu "sin theta))/g`
`=("u"^(2)sin2theta)/(g)(1+e)`
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Knowledge Check

  • A projectile is projected with speed u at an angle theta with the horizontal . The average velocity of the projectile between the instants it crosses the same level is

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    `(u cos theta)/(2)`
    B
    `u cos theta`
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    `u (2 cos (theta)/(2) - sec (theta)/(2))`
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    `u (cos(theta)/(2) - sec(theta)/(2))`

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