Answer
(a) $y_{max} = 81.6~m$
(b) The cart travels 245 meters.
(c) The rocket will land right on the cart.
Work Step by Step
(a) $v_{0y} = 40.0~m/s$
We can find the maximum height $y_{max}$.
$y_{max} = \frac{v_y^2-v_{0y}^2}{2g} = \frac{0-(40.0~m/s)^2}{(2)(-9.80~m/s^2)}$
$y_{max} = 81.6~m$
(b) We can find the time $t$ for the rocket to reach maximum height.
$t = \frac{v_y-v_{0y}}{g} = \frac{0-40.0~m/s}{-9.80~m/s^2}$
$t = 4.08~s$
The total time in the air is $2t$ which is $8.16~s$. We can find the distance $x$ the cart moves in this time.
$x = v~t = (30.0~m/s)(8.16~s) = 245~m$
The cart travels 245 meters.
(c) Since the horizontal speed of the rocket is equal to the horizontal speed of the cart, the rocket will land right on the cart.
