Question

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 up- ward at 4.00 m/s2 until it reaches an altitude of 1 000 m. At that point, its engines fail and the rocket goes into free fall, with an acceleration of 29.80 m/s2. (a) For what time interval is the rocket in motion above the ground?

Answer 1

Answer:

The time interval the rocket is in motion above the ground is the time in the two times the motion is going on $T_{total} = 23,14 s$

Explanation:

$V_{i} = 80 (\frac{m}{s} )$

$S_{i} = 0 m$

The motion in the first step has an acceleration $a_{1} = 4 (\frac{m}{s^{2} } )$

and the maximum height will be and the end of this step is $S_{1} = 1000 m$

So to know the time until the rocket fail and change the acceleration:

$S_{1} = S_{i} + V_{i1} * t + \frac{1}{2} * a_{1} *t^{2}$

$1000m = 0 m + 80 \frac{m}{s} * t + \frac{1}{2} * 4 \frac{m}{s^{2} } * t^{2}$

$1000= 2*t^{2} + 80 * t$ you can divide the expression by two and simplify the calculating

$2t^{2} + 80*t -1000=0$

$t^{2} + 40*t -500=0$

Using quadratic equation :

$\frac{-b +/- \sqrt{b^{2}-4*a*c } }{2*a}$

$\frac{- 40+/- \sqrt{40^{2}-4*-500 } }{2}$

$-20 +/- 30$

$x_{1}= -50 , x_{2}= 10 ,$ The time can be negative so, the time we are going to use is 10s

$t_{1}= 10 s$

Now when the rocket fail it change the direction of the motion and the time is going to be  the time it takes to reach earth again

$v_{f} = v_{i}+a_{1}*t_{1}$

$v_{f}= 80 \frac{m}{s} + 4 \frac{m}{s^{2} } * 10 s = 120 \frac{m}{s}$

$S_{2} = S_{1} + V_{i2} * t + \frac{1}{2} * a_{2} *t^{2}$

$0m = 1000m + 120 \frac{m}{s} * t +\frac{1}{2} (-29,8 \frac{m}{s^{2} })*t^{2}$

$0m = 1000m + 120 \frac{m}{s} * t -14,9 \frac{m}{s^{2} })*t^{2}$

$\frac{-b +/- \sqrt{b^{2}-4*a*c } }{2*a}$

$\frac{- 120+/- \sqrt{120^{2}-4*-14,9*1000 } }{2*14,9}$

$x_{1}= -5,108s , x_{2}= 13,14 ,$ The time can be negative so, the time we are going to use is 13,14s

So the full time is the both times adding them

$T_{total}= 10 + 13,14 = 23,14 s$