A physics Q - Please Help!!~ (with explaination) Thank~

Question

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?

Answer 1

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~~~