So, I have attempted this problem several times, yet I fail to get the correct answer. My question implies that a toy rocket is basically being launched ground up, so y (height) would be 0.00 m and time is at time t = 0.00 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 75 m and acquired a velocity of 10 m/s.The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground with negligible air resistance. The speed of the rocket upon impact on the ground is closest to ...?
Clearly, you can get an intuition that there is no starting velocity, but you are given enough information to solve this. Can you tell me what exact equation I should be using? I may be using the incorrect one.
If you only consider gravity and leave out air resistance (which is physically weird for a rocket, but why not), then this can be elegantly solved (like soooo many physics problems) using conservation of energy.
If you neglect air friction, then the mechanical energy is conserved. At the moment of the burnout, your cinetic energy is $E_c = \frac{m}{2}v_0^2$ where $m$ is the mass of the rocket and $v_0=10\, m\cdot s^{-1}$, and the potential energy is $V=mgh$ where $h=75m$. At landing, the potential energy is $V'=0$ and the cinetic energy is $E'_c = \frac{m}{2}v^2$ where $v$ is your unknown.
So $\frac{m}{2}v_0^2 + mgh = \frac{m}{2}v^2$, and you may divide by $m$ : $\frac{1}{2}v_0^2 + gh = \frac{1}{2}v^2$.
Now you just have to solve numerically to find $v$.