Question

A particle is released from a height S. At certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively:

• A. $\frac { 3 \mathrm { gS } } { 4 }$
• B. ,
• C. ,
• D. ,

Answer 1

Answer: (D)

,

Answer 2

T.E_i = T.E_f

mgS + 0 = 1/2 mv² + mgh … (1)

and 1/2 mv² = 3mgh … (2)

Solve (1) and (2) we get

h = S/4 and v = $\sqrt { \frac { 3 \mathrm { gS } } { 2 } }$