Physics Question (Energy Change)?

Question

I'm doing some review, and I stumbled upon this practice question: 

A rocket is fired upwards at 12 ms-1

How high does it go?
What is it's velocity at 3 m above the ground? 


*I have no mass, no time, and no other information provided. How do I solve it?

Rich · Accepted Answer

Mass is not a big deal.  Use Vf^2 = Vi^2 + 2ax

Vf in the first part is 0 m/s because it is at the top.  Plugging everything in (a=-9.81m/s^2), gives you 7.3m as the maximum height.

In the second part, you are looking for x.  Doing all the math (don't forget that it's vf^2!), you get 9.2 m/s as your Vf.

Anonym · Answer

y = height Max = u * t-1/2 g * t ^ 2
u=12 m/s,
t = u / g
t = 12m / s: 9.81m / s ^ 2 = 1.22 s
y = 12m / s * 1.22s - 0.5 * 9.81m / s ^ 2 * (1.22s) ^ 2
y = 7.34m

v (above 3m )=  u-â2hg
v = 12 m / s - â 2 * 3m * 9.81m / s ^ 2 =
12m/s - 7.67 m / s =
v = 4.33m / s

v^2 = u^2 - 2as ===> u -â2as  a=g; h=s

Physics Question (Energy Change)?

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