A test rocket is fired vertically upward from a well. A catapultgives it an init

Question

A test rocket is fired vertically upward from a well. A catapultgives it an initial speed of 80.4 m/s atground level. Its engines then fire and it accelerates upward at4.10 m/s2 until it reaches analtitude of 940 m. At that point itsengines fail, and the rocket goes into free fall, with anacceleration of -9.80 m/s2. (You will need to considerthe motion while the engine is operating separate from thefree-fall motion.) (a) How long is the rocket in motion above theground?
Enter anumber. s

(b) What is its maximum altitude?
Enter anumber. km

(c) What is its velocity just before it collides with theEarth?
Enter anumber. m/s I am new to physics and am not sure how to dothis one.your help would be appreciated (a) How long is the rocket in motion above theground?
Enter anumber. s

(b) What is its maximum altitude?
Enter anumber. km

(c) What is its velocity just before it collides with theEarth?
Enter anumber. m/s Enter anumber. Enter anumber. Enter anumber. I am new to physics and am not sure how to dothis one.your help would be appreciated

Explanation / Answer

Initial speed v = 80.4 m / s Accleration a = 4.1 m / s ^ 2 Altitude h = 940 m Speed at height h be V then from the relation V ^ 2 - v ^ 2 = 2ah                                           V = [ v ^ 2 + 2ah ]                                              = 119 m / s height reached by rocket above height h is h ' = V ^ 2 /2g                                                                     = 723 m So. Maximum height reached by the rocket from the ground H = h + h '                                                                                               = 1663 m timetaken to reach the height h is t = ( V - v ) /a                                                            = 9.414s time taken to reach H from height h is t ' = V /g                                                               = 12.142 s So, total time to reach height H from ground T = t + t'                                                                       = 21.55 s time taken to reach the ground from height H is t " = [2H / g ]                                                                                    = 18.422 s So. total time in air = t + t ' + t "                               = 39.97s (b). maximum height H = 1663 m (c). velocity just before hits the ground = [ 2gH]                                                            =180.54 m / s

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