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A stone is released from the top of a tower of height 16.6 m calculate it's flnal velocity before touching the ground. |
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Answer» nitial Velocity u=0Initial Velocity u=0Fianl velocity V=?Initial Velocity u=0Fianl velocity v=?HEIGHT, s=19.6mInitial Velocity u=0Fianl velocity v=?Height, s=19.6mBy third EQUATION of motionInitial Velocity u=0Fianl velocity v=?Height, s=19.6mBy third equation of motionv Initial Velocity u=0Fianl velocity v=?Height, s=19.6mBy third equation of motionv 2Initial Velocity u=0Fianl velocity v=?Height, s=19.6mBy third equation of motionv 2 =u Initial Velocity u=0Fianl velocity v=?Height, s=19.6mBy third equation of motionv 2 =u 2Initial Velocity u=0Fianl velocity v=?Height, s=19.6mBy third equation of motionv 2 =u 2 +2gsInitial Velocity u=0Fianl velocity v=?Height, s=19.6mBy third equation of motionv 2 =u 2 +2gsv Initial Velocity u=0Fianl velocity v=?Height, s=19.6mBy third equation of motionv 2 =u 2 +2gsv 2Initial Velocity u=0Fianl velocity v=?Height, s=19.6mBy third equation of motionv 2 =u 2 +2gsv 2 =0+2×9.8×19.6Initial Velocity u=0Fianl velocity v=?Height, s=19.6mBy third equation of motionv 2 =u 2 +2gsv 2 =0+2×9.8×19.6v Initial Velocity u=0Fianl velocity v=?Height, s=19.6mBy third equation of motionv 2 =u 2 +2gsv 2 =0+2×9.8×19.6v 2Initial Velocity u=0Fianl velocity v=?Height, s=19.6mBy third equation of motionv 2 =u 2 +2gsv 2 =0+2×9.8×19.6v 2 =384.16Initial Velocity u=0Fianl velocity v=?Height, s=19.6mBy third equation of motionv 2 =u 2 +2gsv 2 =0+2×9.8×19.6v 2 =384.16⇒v=19.6m/sMark me as Brainlist |
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