Question

A monkey is launched straight up in a rocket w/t acceleration of 12.0m/s^2 (up). When the rocket's velocity is 450.0m/s (up), the engine shuts off. What is the rockets height from the ground when its velocity has dropped to 200m/s (U). step by step solution.

Answer 1

can you answer it?! @carson889

Answer 2

Total distance: s = s_1 + s_2

Distance traveled under power:
u_1 = initial velocity = 0 m/s
\[v_{1} ^{2}=u_{1} ^{2}+2*a_{1}*s_{1}\]
(450 m/s)^2 = 2*(12 m/s^2)*s_1
\[s _{1}= \frac{ 450^{2} }{ 24 } = 8437.5\]


a_2 = Gravity = -9.8 m/s^2 [up]
u_2 = v_1 = 450 m/s
\[v_{2} ^{2}=u_{2} ^{2}+2*a_{2}*s_{2}\]
(200 m/s)^2 = (450m/s)^2 + 2*(-9.8 m/s^2)*s_2
\[s_{2}=\frac{ 200^{2}-450^{2} }{ 2*-9.8 } =8290.8 \]

s = s_1 + s_2 = 8437.5m + 8290.8m = 16728.3 m

Therefore, the rocket is 16728.3 meters from the ground when the velocity drops to 200 m/s.

Answer 3

Thank =)

Answer 4

Your welcome!