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A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.0 m/s at ground level. The engines then fire, and the rocket accelerates upward at 4.00 m/s\(^2\) until it reaches an altitude of 1000 m. At that point, its engines fail and the rocket goes into free fall, with an acceleration of 9.80 m/s\(^2\).

(a) For what time interval is the rocket in motion above the ground?

Answer :

Answer:

There is an interval of 24.28s in which the rocket is above the ground.

Explanation:

[tex]y_{i}=0m[/tex]

[tex]v_i=80\frac{m}{s}[/tex]

[tex]a=4\frac{m}{s^2}[/tex]

[tex]y_{e}=1000m[/tex]

[tex]g=9.8\frac{m}{s^2}[/tex]

From Kinematics, the position [tex]y[/tex] as a function of time when the engine still works will be:

[tex]y(t)=v_it+\frac{1}{2}at^2[/tex]

At what time the altitud will be [tex]y_{e}=1000m[/tex]?

[tex]v_it+\frac{1}{2}at^2=y_{e}[/tex] ⇒ [tex]\frac{1}{2}at^2+v_it-y_{e}=0[/tex]

Using the quadratic formula: [tex]t_1=10s[/tex].

How much time does it take for the rocket to touch the ground? No the function of position is:

[tex]y(t)=y_{e}+v_et-\frac{1}{2}gt^2[/tex]

Where our new initial position is [tex]y_{max}[/tex], the velocity when the engine breaks is [tex]v_e=v_i+at=120\frac{m}{s}[/tex] and the only acceleration comes from gravity (which points down).

Now, when the rocket tounches the ground:

[tex]y_{e}+v_et-\frac{1}{2}gt^2=0[/tex]

Again, using the quadratic ecuation:

[tex]t_2=24.49s[/tex]

Now, the total time from the moment it takes off and the moment it tounches the ground will be:

[tex]t_T=t_1+t_2=34.49s[/tex].

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Rewritten by : Barada