A model rocket fired vertically from the ground ascends with avertical accelerat

Question

A model rocket fired vertically from the ground ascends with avertical acceleration of 3.6m/s2 for 5.0 s. Its fuel isthen exhausted, so it continues upward as a free-fall particle andthen falls back down. (a) What is the maximum altitude reached?
(b) What is the total time elapsed from takeoff until therocket strikes the ground?
A model rocket fired vertically from the ground ascends with avertical acceleration of 3.6m/s2 for 5.0 s. Its fuel isthen exhausted, so it continues upward as a free-fall particle andthen falls back down. (a) What is the maximum altitude reached?
(b) What is the total time elapsed from takeoff until therocket strikes the ground?

Explanation / Answer

Lets do this problem in 2 parts First part is during the 5 sec of 3.6m/s acceleration after the 5 seconds, the speed is 5sec * 3.6 = 18m/s distance = 1/2*a*t^2 = .5*3.6*25 = 45m so after the 5 seconds, the rocket is 45 m in the air with a speedof 18 m/s to hit the ground, the rocket must fall 45m d=vo*t + 1/2*a*t^2 -45=18m/s*t + 1.2 * -9.81m/s^2 * t^2 using the quadratic formula, t=5.38 sec therefore, the total time for rocket to hit the ground is 5.38+5 =10.38 sec find distance where v2=0 to find max height v2^2=v1^2 + 2 *a *d 0=18m/s^2 + 2*-9.81*d, d=4.128m total distance = 45m + 4.128m = 49.128m

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