A spherical ball of mass 20 kg is stationary at the top of the hill of height 100 m . It rolls down smooth surface to ground then climb up another hill of height 40 m and finally rolls down to a horizontal base at a height of 15 m above the ground . The maximum velocity attained by a ball is `....... ( g = 9.8 m//s^(2))`

Options

A. `40 ms^(-1)`
B. `40 . 8 ms^(-1)`
C. `34.3 ms^(-1)`
D. None of these

Answer

B

Explanation

`1/2 mv_(1)^(2) +mgh_(1)=1/2 mv_(2)^(2) +mgh_(2)`
`:. Mgh_(1)-mgh_(2)=1/2mv_(2)^(2) -1/2mv_(1)^(2)`
` :. Mg (h_(1) - h_(2)) = 1/2 mv_(2)^(2) " " [ :. v_(1) = 0 ] `
` :. 20 xx 9.8 [100-15] = 1/2 xx2v xxv_(2)^(2)`
` :. (20 xx 9.8 xx85 xx2)/20 = v_(2)^(2)`
` :. 40.816 ms^(-1) = v_(2)`
` :. v_(2) = 40.8 ms^(-1)`