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A ball is projected with velocity 20sqrt...

A ball is projected with velocity `20sqrt(2)` m/sec at an angle of `45^(@)` with horizontal towards a fixed solid hemi-cylinder as shown in figure such that it collide with the cylinder at its highest point of trajectory. If the collision is perfectly elastic. Then (take `sqrt(3)=1.7` and `sqrt(7)=2.7, g=10m//s^(2)`)

A

maximum height attained by the ball from the horizontal surface is 35 m before collision with the horizontal surface.

B

time from t = 0, when it collide with the ground is 6.4 sec (approximately).

C

after the collision with the solid hemi0cylinder, ball again collides with the solid hemi - cylinder.

D

the position where it collide on the ground first time from the point of projection is 4 m (approximately).

Text Solution

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The correct Answer is:
A, B, D
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Knowledge Check

  • A particle is projected with velocity of 10m/s at an angle of 15° with horizontal.The horizontal range will be (g = 10m/s^2)

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    `20sqrt5 m//s`
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