A physics Q - Please Help!!~ (with explaination) Thank~
A water rocket moves vertically upward from rest with an acceleration of 15 ms-2 when water is being ejected. The ejecting of water lasts for 2 s. It then moves freely under gravity. What is the maximum heigth reached by the water rocket?
- Audrey HepburnLv 71 十年前最愛解答
Take upward direction be positive.
Then, find out the height gained by the rocket in the first 2 seconds
By s = ut + 1/2 at2
s = (0)(2) + 1/2 (15)(2)2
s = 30 m
Therefore, the height gained by the rocket in the first 2 seconds is 30 m.
By v = u + at
v = (0) + 15(2)
v = 30 ms-1
So, the velocity of the rockect is 30 ms-1 upward.
After 2 seconds, the rocket experiences a downward force due to its weight. That is, it will acclerate downward by free falling.
At maximum height, the instantaneous velocity of the rocket is 0.
So, by v2 = u2 + 2as
(0)2 = (30)2 + 2(-10)s
s = 45 m
So, after the ejecting of water, the rocket can still go up by 45 m.
As a result, maximum height gained by the water rocket
= 30 + 45
= 75 m資料來源： Myself~~~