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The coach throws a baseball to a player ...

The coach throws a baseball to a player with an initial speed at `20ms^(-1)` at an angle of `45^(@)` with the horizontal.At the moment the ball is thrown, the player is `50m` from coach.The speed and the direction that the player has to run to catch the ball at the same height at which it was released in `ms^(-1)` is

A

`5/sqrt2`away from coach

B

`5/sqrt2` towards from coach

C

`sqrt2/5`towards the coach

D

`sqrt2/5`away from the coach

Text Solution

Verified by Experts

The correct Answer is:
B
`R_("ball")=(u^(2)sin2theta)/g,s=vt`
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The coach throws a baseball to a player with an initial speed of 20 m//s at an angle of 45^@ with the horizontal. At the moment the ball is thrown, the player is 50 m from the coach. At what speed and in what direction must the player run to catch the ball at the same height at which it was released? (g = 10 m//s^2) .

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Knowledge Check

  • The maximum height attained by a ball projected with speed 20 ms^(-1) at an angle 45^(@) with the horizontal is [take g = 10 ms^(-2) ]

    A
    40 m
    B
    20 m
    C
    10 m
    D
    30 m

  • A cricket ball is thrown at a speed of 30 m s^(-1) in a direction 30^(@) above the horizontal. The time taken by the ball to return to the same level is

    A
    2 s
    B
    3 s
    C
    4 s
    D
    5 s

  • A ball A is projected with speed 20 m//s at an angle of 30^(@) with the horizontal from the ground. Another ball B is released simultaneously from a point on the vertical line along the maximum height of the projectile. Both the balls collide at the maximum height of the first ball. The initial height of ball B is

    A
    `5 m`
    B
    `10 m`
    C
    `15 m`
    D
    `20 m`

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