Calculating Rocket Acceleration for a 200 m/s Speed at 1.0km Height

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To calculate the acceleration needed for a rocket to reach a speed of 200 m/s at a height of 1.0 km, the relevant formula is v_f^2 = v_i^2 + 2ad. In this scenario, the initial velocity (v_i) is 0 m/s, the final velocity (v_f) is 200 m/s, and the displacement (d) is 1000 m. By rearranging the formula, acceleration (a) can be determined. The discussion clarifies that while calculus is often involved in advanced physics, basic kinematic equations can suffice for this calculation. Understanding these variables is crucial for solving the problem effectively.
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What acceleration does a rocket need to reach speed of 200 m/s at height of 1.0km. I know that acceleration is velocity over time but I have only velocity and distance. :confused:
 
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So how about the formula:
v_f ^ 2 = v_i ^ 2 + 2ad? Do you know this formula?
Can you go from there? Remember to change 1.0 km into 1000m.
Viet Dao,
 
That is calculus. Would you mind telling me those variables.
 
a : acceleration (m / s ^ 2) = ?.
d : displacement (m) = 1000 m.
v_i : initial velocity (m / s) = 0 m / s (The object accelerates from rest).
v_f : final velocity (m / s) = 200 m / s.
Can you go from here?
By the way, I think it is physics, not calculus...
Viet Dao,
 
O thanks. I tried to look it up in some later chapters but found formulas involving derivations. I didn't know what those variables were so I guessed that you have to use calculus to find way to do it. Thanks for help. :smile:
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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