A rocket is launched straight up with an initial velocity of 40 m/s. After 3 seconds, its engine shuts off, and it continues moving upwards until gravity stops it momentarily before it begins to fall back to the ground. What is the maximum height the rocket reaches?, how long does it take for the rocket to reach the ground after the engine shutdown?, what is the total time the rocket spends in the air?
Answer (1)
The rocket starts off traveling 40 m/s for 3 seconds. The height achieved in this period of time is determined by multiplying the velocity by time. Thus, at this stage, the height is 120m.When the engine shuts off, the rocket does not stop immediately, rather it continues going up until Earth's gravity brings it to a stop. To get the maximum height, use one of the kinematic equation in the form of[tex]v_f^{2} = v_i^{2} +2ax[/tex]Substituting known values, [tex]0 = 40^{2} + 2(-9.8)x[/tex], x is determined to be 81.63m. (The value of a is negative since it still opposes gravity's acceleration).We can also determine the time the rocket took before it achieved the additional 81.63m, this time by using[tex]v_f = v_i + at[/tex]Substituting known values, [tex]0 = 40-9.8t[/tex], t = 4.08sThe maximum height is obtained by adding the height achieved when the rocket is still working and when it shuts down.max h = 120 + 81.63 = 201.63mNext, to get the time the rocket took from engine shutdown to reaching the ground, we need to obtain the time the rocket took from max height to the ground. Using the kinematic equation[tex]x = v_it + \frac{1}{2}at^{2}[/tex]Substituting known values, [tex]201.63 = 0(t) + \frac{1}{2}(9.8)(t)^{2}[/tex], t = 6.41 s.Thus, from engine shutdown to touching the ground, the total time ist = 6.41 + 4.08 = 10.49 secondsTo get the total time the rocket spends in the air, add the times from initial lift-off to it going back to the groundt (total) = 6.41 + 4.08 + 3 = 13.49 seconds
