Phy Question Projectile motion

2012-06-10 6:49 am

1.a stone is thrown upwards from the top of a cliff with a initial velocity 10ms^-1at an elevation of 30°.The height of the cliff is 30m. After time t,the stone strikes the ground at point P .

(a)Find t
(b)the distance R od point P from he foot of the cliff
(c)the volecity of the stone at point P
(d) the greatest height H above reached bt the stone

a) I know it is

s= ut + (1/2)at^2
where s = -30m, u=10sin30* = 5ms^-1, a=-g=-10

but why the height is -30m,

the time for the ball to move to the height at 30m,
is equal to the ball is thrown from a cliff with height 30m to the ground?

Always like this?

回答 (1)

天同

2012-06-10 7:43 am

✔ 最佳答案

Q: why the height is -30m,

A: -30 m is the displacement of the stone.
In the equation s = ut + (1/2)at^2 , s is the displacement, which is defined as the final position minus the initial position.

Now, if you take the ground as the reference level, the initial position of the stone is +30 m (i.e. 30 m above the ground), whereas the final position is 0 m (ground level). Hence,

displacement of the stone = [0 - (30)] m = -30 m

2012-06-11 23:33:16 補充：
Answer to your 1st suppl question: Yes, you are right.

2012-06-11 23:36:37 補充：
2nd question:
The displacements for the upward journey from cliff to the sky (say, +h) and the downward journey from the sky back to the cliff (say, -h) canceled out one another (+h+(-h) = 0). The resultant displacement is thus zero.

2012-06-11 23:37:57 補充：
(cont'd)...Thus you only need to calcuate the displacement from cliff to ground.

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