free fall motion question

用戶100130

2007-10-28 12:24 am

A boy jump and his feet are 0.5m above the ground.
find the hang-time of the boy.
solution:
s=ut=1/2at^2
0.5=1/2(10)t^2
t=0.316s
hang time=0.316x2
=0.632s
問題:
點解要計完jump上去的time taken, 然後再x2,計成個action 的time taken。
可以一次過計個hang time嗎﹖點解﹖

回答 (1)

Gabriella Montez

2007-10-28 5:03 am

✔ 最佳答案

Take downwards as +VE.

The above calculations consider the downwards journey where the boy returns to the ground from the highest point ( 0.5m above ground ), so u = 0 , a = 10, s = 0.5:

s = ut + at^2 / 2

0.5 = ( 0 )( t ) + ( 10 )( t^2 ) / 2

t = 0.316s

But it's only half of the journey, the time taken for the upwards journey ( another 0.316s ) has not been taken in account.

Therefore the hang-time = 0.316 x 2 = 0.632s

= = = = = =

Alternative method: consider the upwards journey

( v = 0, a = 10, s = -0.5 )

v^2 - u^2 = 2as

0^2 - u^2 = 2 ( 10 )( - 0.5 )

u = -3.16m/s ( since downwards is taken as +VE )

As the boy returns to his starting point, s = 0 in the whole journey,

s = ut + at^2 / 2

0 = ( - 3.16 )( t ) + ( 10 )( t^2 / 2 )

t = 0 ( rejected ) or t = 0.632s which is the hang-time

參考: My Phy Knowledge

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