A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25m/s2 and feels no appreciable air resistance. When it has reached a height of 525 m, its engines suddenly fail; the only force acting on it is now gravity. (a) What is the maximum height this rocket will reach above the launch pad? (b) How much time will elapse after engine failure before the rocket comes crashing down to the launch pad, and how fast will it be moving just before it crashes? (c) Sketch ay−t, vy−t, and y - t graphs of the rocket’s motion from the instant of blast-off to the instant just before it strikes the launch pad.

H = 645.42 m

v_4 = 112.53 m/s

T = 16.43 s

Explanation:

Given:-

- The mass of the rocket m = 7500 kg

- The acceleration for first part of the journey a_1 = 2.25 m/s^2

- The height reached before engine failure h_1 = 525 m

Find:-

(a) What is the maximum height this rocket will reach above the launch pad? (b) How much time will elapse after engine failure before the rocket comes crashing down to the launch pad, and how fast will it be moving just before it crashes?

(c) Sketch ay−t,vy−t, and y - t graphs of the rocket’s motion from the instant of blast-off to the instant just before it strikes the launch pad.

Solution:

- Find the velocity v_2 when the engine fails. Using 3rd equation of motion.

                          v_2^2 = v_i^2 + 2*a_1*h_1

The rocket blasts off from rest v_i = 0.

                          v_2 = sqrt ( 2*a_1*h_1 )

                          v_2 = sqrt ( 2*2.25*525 ) = sqrt ( 2362.5 )

                          v_2 = 48.606 m/s

- The time taken for first part of upward journey t_1. Using first equation of motion:

                          v_2 = v_1 + a_1*t_1

                          t_1 = v_2 / a_1

                          t_1 = 48.606 / 2.25

                          t_1 = 21.603 s            

- Now for second part of the upward journey, the initial velocity is v_2 and final velocity is when rocket reaches maximum height v_3 = 0. The only force is gravity; hence, a_2 = g = -9.81 m/s^2. Using 3rd equation of motion.

                          v_3^2 - v_2^2 = 2*a_2*h_2

                          h_2 = v_2^2 / 2*g

                          h_2 = 2362.5 / 2*9.81 = 120.415 m

- The maximum height the rocket will reach above launch pad is H:

                         H = h_1 + h_2

                         H = 525 + 120.415

                         H = 645.42 m

- The time taken t_2 for second part of upward journey, Using first equation of motion is:

                        v_3 = v_2 + a_2*t_2

                        t_2 = v_2 / g

                        t_2 = 48.606 / 9.81

                        t_2 = 4.955 s

- The final velocity v_4 as soon as the rocket crashes down. Using 3rd equation of motion.

                       v_4^2 = v_3^2 + 2*a_2*H

                       v_4 = sqrt ( 2*a_2*H )

                       v_4 = sqrt ( 2*9.81*645.42 ) = sqrt ( 12663.1404 )

                       v_4 = 112.53 m/s

- The time taken t_3 for downward part of journey, Using first equation of motion is:

                        v_4 = v_3 + a_2*t_3

                        t_3 = v_4 / g

                        t_3 = 112.53 / 9.81

                        t_3 = 11.47 s

- The total Time taken after engine failure to crash is T:

                        T = t_2 + t_3

                        T = 11.47 + 4.955

                        T = 16.43 s

the temperature and the pressure affect solubility. temperature affects by changing the properties of the solvent. the increase in pressure increases solubility, the decrease in pressure decreases solubility.

massxgravity x heght=weightxheight= pe

4900=490xh

h=10