a rocket takes off vertically from rest under the power of its rocket motor. The rocket has has an upward acceleration of 5g. after 5 sec the motor instantaneou

Question

Please show work. I really need help figuring out this problem.

 

a rocket takes off vertically from rest under the power of its rocket motor. The rocket has has an upward acceleration of 5g. after 5 sec the motor instantaneously burns out and only the downward acceleration of gravity acts on the rocket. Approximately what maximum height does the rocket reach above the point where its motor burns out

Answer 1

First, find the upward vertical speed of the rocket when the motor burns out.

 

v(t) = v₀ + at = 0 + (5g)t = 5gt

 

v(5) = [5(9.8)](5) m/s = 245 m/s

 

Now place the origin at the height where it burns out, and solve the rest of the problem using kinematics.  In this part of the problem, the acceleration is -g, the initial speed is the result for v(5) from above, and the final speed is zero (comes to instantaneous rest at the top of the trajectory).

 

v² - v₀² = 2ah

 

0² - v(5)² = 2(-g)h

 

h = v(5)²/(2g) = (245)² / [2(9.8)] m = 3062.5 m