How High Does a Rocket Go If It Accelerates for 4 Seconds?

[bkoz316]

Finding Maximum Height?

A rocket moves striaght upward, starting from rest with an acceleration of 29.4m/s^2. It runs out of fuel at the end of 4.00seconds and continues to coast upward, reaching a maximum height before falling back to Earth.

(a.) Find the rocket's velocity and position at the end of 4.00seconds.

(b.) Find the maximum height the rocket reaches.






There was ten parts to this problem and I'm stuck on how todo these to.
Thanks sooo much for the help!

 

[dzogi]

a) $$s=\frac{1}{2}at^{2}; v=at$$
b) at the end of the 4 seconds period, the rocket will have velocity of $$v$$ (calculated above) and a downward negative acceleration of $$g=-9.81m/s^{2}$$. Do the rest yourself :)

Hint: You can also use the law of conversion of energy for the second part.

 

[ogb p]

I would approach this problem by first understanding the basic principles of motion and applying them to the given scenario. The rocket's motion can be described using the equations of motion, specifically the equations for displacement, velocity, and acceleration.

(a.) To find the rocket's velocity and position at the end of 4.00 seconds, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity (in this case, 0 m/s), a is the acceleration (29.4 m/s^2), and t is the time (4.00 seconds). Plugging in the values, we get v = 0 + (29.4)(4.00) = 117.6 m/s. This is the rocket's velocity at the end of 4.00 seconds.

To find the position, we use the equation s = ut + 1/2at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. Again, plugging in the values, we get s = (0)(4.00) + 1/2(29.4)(4.00)^2 = 235.2 m. This is the rocket's position at the end of 4.00 seconds.

(b.) To find the maximum height the rocket reaches, we can use the equation v^2 = u^2 + 2as, where v is the final velocity (which is 0 m/s at the maximum height), u is the initial velocity, a is the acceleration, and s is the displacement (which is the maximum height we are trying to find). Rearranging the equation, we get s = (v^2 - u^2)/2a. Plugging in the values, we get s = (0 - 117.6^2)/2(-9.8) = 725.28 m. This is the maximum height the rocket reaches before falling back to Earth.

In summary, using the equations of motion, we can find the rocket's velocity and position at the end of 4.00 seconds and also calculate the maximum height it reaches before falling back to Earth. These calculations rely on the assumption that the rocket's acceleration remains constant throughout its motion. However, in reality, factors such as air resistance may affect the rocket's motion and the calculations may not be entirely