Question

two rockets sit at rest on a launch - pad. the rocket on left is capable of accelerating upwards at a rate of 15 m/s/s, while the rocket on the right can accelerate upwards at a greater rate of 30 m/s/s. lets assume both acceleration values are uniform. the rocket on the left is launched directly upwards first, then 20 seconds later the rocket on the right is also launched directly upwards.

• how long after being launched will the rocket on the right reach the same height above the ground as the rocket on the left? (10pts)
• what will be the height of the two rockets when they are at the same height above the ground? (5pts)
• what will be the velocities of the two rockets when they are the same height above the ground? (5pts)

Explanation:

Step1: Write the displacement formula for the left - hand rocket

The displacement of an object under constant acceleration starting from rest ($u = 0$) is given by $s=ut+\frac{1}{2}at^{2}$. For the left - hand rocket, $a_1 = 15\ m/s^{2}$ and it starts at $t = 0$. So, $s_1=\frac{1}{2}a_1t^{2}=\frac{1}{2}\times15t^{2}=7.5t^{2}$.

Step2: Write the displacement formula for the right - hand rocket

The right - hand rocket starts 20 s later. Let the time for the right - hand rocket be $t'=t - 20$. Its acceleration $a_2 = 30\ m/s^{2}$ and starting from rest ($u = 0$). So, $s_2=\frac{1}{2}a_2(t - 20)^{2}=\frac{1}{2}\times30(t - 20)^{2}=15(t - 20)^{2}$.

Step3: Set $s_1=s_2$ to find the time when they are at the same height

$7.5t^{2}=15(t - 20)^{2}$
$7.5t^{2}=15(t^{2}-40t + 400)$
$7.5t^{2}=15t^{2}-600t + 6000$
$7.5t^{2}-15t^{2}+600t - 6000 = 0$
$- 7.5t^{2}+600t - 6000 = 0$
Multiply through by - 1 to get $7.5t^{2}-600t + 6000 = 0$. Divide by 7.5: $t^{2}-80t + 800 = 0$.
Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 1$, $b=-80$ and $c = 800$.
$t=\frac{80\pm\sqrt{(-80)^{2}-4\times1\times800}}{2\times1}=\frac{80\pm\sqrt{6400 - 3200}}{2}=\frac{80\pm\sqrt{3200}}{2}=\frac{80\pm40\sqrt{2}}{2}=40\pm20\sqrt{2}$.
We take the positive root $t = 40 + 20\sqrt{2}\approx40+20\times1.414=40 + 28.28=68.28\ s$.

Step4: Find the height when they are at the same height

Substitute $t = 40 + 20\sqrt{2}$ into $s_1$:
$s_1=7.5t^{2}=7.5\times(40 + 20\sqrt{2})^{2}=7.5\times(1600+1600\sqrt{2}+800)=7.5\times(2400 + 1600\sqrt{2})=18000+12000\sqrt{2}\approx18000+12000\times1.414=18000 + 16968=34968\ m$.

Step5: Find the velocities of the two rockets

The velocity formula is $v = u+at$.
For the left - hand rocket: $v_1=a_1t=15\times(40 + 20\sqrt{2})=600+300\sqrt{2}\approx600 + 300\times1.414=600+424.2 = 1024.2\ m/s$.
For the right - hand rocket: $v_2=a_2(t - 20)=30\times(40 + 20\sqrt{2}-20)=30\times(20 + 20\sqrt{2})=600+600\sqrt{2}\approx600+600\times1.414=600 + 848.4=1448.4\ m/s$.

Answer:

• The rocket on the right reaches the same height as the rocket on the left $40 + 20\sqrt{2}\approx68.28\ s$ after being launched.
• The height of the two rockets when they are at the same height is $18000+12000\sqrt{2}\approx34968\ m$.
• The velocity of the rocket on the left is $600+300\sqrt{2}\approx1024.2\ m/s$ and the velocity of the rocket on the right is $600+600\sqrt{2}\approx1448.4\ m/s$.