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A stone is thrown vertically upward with...

A stone is thrown vertically upward with a speed of 28 m/s. (a) Find the maximum height reached by the stone. (b)Find its velocity one second before it reaches the maximum height.

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The correct Answer is:
C, D
Given `u=28 m/s, v=0, a =-g=-9.8 m/s^2, S_2 = (v^2-u^2)/(2a), =(0^2-28^2)/(2(-9.8)=2.85, t'=2.85-1=1=1.85, v'=u+at'=28-(9.8)(1.85), (28-18.13=9.87m/s` Hence the velocity is 9.87m/s. c. No it will not change. As, after one second velocity becomes zero for any initial velocity and deceleration is `g= 9.8 m/s^2` remains same. for initial velocity more than 28 m/s maximum height increases.
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Knowledge Check

  • A particle is thrown vertically upwards its velocity at half of the height is 10 m/s then the maximum height attained by it is ( g= 10 m//s^2)

    A
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    B
    `20 m`
    C
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    D
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