physic 拋物向上求building height
係天台拋一樣野向上 with speed 16ms-1
個樣野會reach the ground after 4seconds
find the height of the building
the Answer is 16 m !!!!!
agnissammy and seiya
- Gabriella MontezLv 71 十年前最愛解答
Divide the journey into two main parts. Take downwards as positive.
In the first part, the object reaches its maximum height and then returns to original position, which is at the top of the building.
v2 – u2 = 2as
( 16 )2 – ( 0 )2 = 2 ( 10 )( s )
s = 12.8 m
s = ut + at2 / 2
12.8 = ( 0 )( t ) + ( 10 )( t2 ) / 2
t = 1.6s
But it is only half of the first part of the journey; hence the time for the 1st part is 3.2s.
Then the time for the second part of the journey will be 4 – 3.2 = 0.8s and at the start of the second part, the object has an initial v of 16m/s.
s = ut + at2/ 2
s = ( 16 )( 0.8 ) + ( 10 )( 0.8 )2 / 2
= 16 m
So the height of the building is 16m.資料來源： My Phy Knowledge
- 志仁Lv 61 十年前
s=ut+1/2 a t^2
The building is 30.4m in height.
The minus sign mean the ground is under your feet at 30.4m.
2007-07-25 12:52:12 補充：
- 1 十年前
s = ut+ 1/2 (at^2)
2007-07-25 12:50:22 補充：
s = height of the buildingu= initial speeda= gravitational accelerationt=time taken