Answer to Question #162515 in Physics for Chrishan

Question

 A model rocket is released straight upward with an initial speed of 60.5 m/s. It accelerates with a 

constant upward acceleration of 2.50 m/s2 until its engines stop at an altitude of 175. m. (a) What can you 

say about the motion of the rocket after its engines stop? (b) What is the maximum height reached by the 

rocket? (c) How long after liftoff does the rocket reach its maximum height? (d) How long is the rocket in 

the air? 

Answer 1

a) It is free falling under gravity force.

b)

$$V^2=v^2+2ah\\V^2=60.5^2+2(2.5)(175)\\V=67.3\frac{m}{s}$$

$$H=h+\frac{V^2}{2g}\\H=175+\frac{67.3^2}{2\cdot9.8}=406\ m$$

c)

$$t_1=\frac{67.3-60.5}{2.5}=2.72\ s$$

$$t_2=2.72+\frac{67.3}{9.8}=9.59\ s$$

d)

$$t_3=\sqrt{\frac{2(406)}{9.8}}=9.10\ s$$

$$T=9.1+9.6=18.7\ s$$