Question

A stone is released from the top of a tower of height 16.6 m calculate it's flnal velocity before touching the ground. ​

Answer 1

ANSWER

Initial Velocity u=0

Initial Velocity u=0Fianl velocity v=?

Initial Velocity u=0Fianl velocity v=?Height, s=19.6m

Initial Velocity u=0Fianl velocity v=?Height, s=19.6mBy third equation of motion

Initial 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 2

Initial 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 2

Initial Velocity u=0Fianl velocity v=?Height, s=19.6mBy third equation of motionv 2 =u 2 +2gs

Initial 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 2

Initial Velocity u=0Fianl velocity v=?Height, s=19.6mBy third equation of motionv 2 =u 2 +2gsv 2 =0+2×9.8×19.6

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

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 2

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 2 =384.16

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 2 =384.16⇒v=19.6m/s

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Answer 2

Answer:

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