The rocket's acceleration has components \(a_{x}(t)= \alpha t^{2}\) and \(a_{y}(t)= \beta - \gamma t\), where \(\alpha = 2.50 {\rm m}/{\rm s}^{4}, \beta = 9.00 {\rm m}/{\rm s}^{2}\), and \(\gamma = 1.40 {\rm m}/{\rm s}^{3} \). at \(t = 0\) the rocket is at the origin and has velocity \({\vec{v}}_{0} = {v}_{0x} \hat{ i } + v_{0y} \hat{ j }\) with \(v_{0x} = 1.00 {\rm m}/{\rm s}\) and \(v_{0y} = 7.00 {\rm m}/{\rm s}\)

what is the max height? ?

 it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt 
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x} 
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y} 
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ] 
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt 
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases 
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume 
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ] 

Sorry my friend i don't know the answer. i am just in 6th

Electoral current force mechanical advantage if the wheel and axle is 2 n and the output force is 20 n