dynamics in two dimensions

[quick]

A 500 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exerts an 8.0 N thrust on the rocket. Your goal is to have the rocket pass through a small horizontal hoop that is 20 m above the launch point.

i already have the answer of 7.62 but i need to know how to do it. could someone explain how to get the solution?

[quick]

i realize that the acceleration for the x component is 0. the acceleration in y component is just 8 / .5 = 16 m/s^2
the position at time t is
x = v_0*t = 3.0t
y = .5* A_y*t^2 = 8*t^2
im trying to get it to pass a hoop that is 20 m high so i tried 20 = 8*t^2 and i got a time of 1.58 s. i tried putting that time back into the x position equation but i only get 4.74. what am i doing wrong?

[vsage]

What is the question asking for? You just sort of gave us a situation and a goal but not what unknown conditions we need to solve for. Are you asking for at what distance away from the hoop horizontally the thrusters should be turned on at?

Why would 20 = 8*t^2? if 8 is thrust in N then your units are like not position but rather mass * position. The vertical component of displacement is given by d = 1/2*a*t^2. If 8 is the force what is the acceleration?

[quick]

thats exactly what i want to find out. but i already know the distance, i need to find how to get that answer.

[vsage]

(read above post)

[quick]

ok i got acceleration by force/mass = 8 / .5 = 16 m/s^2
and then using the equation y = y_0 + v_0(t - t_0) + 1/2(a(t-t_0)^2
so i set 20 as y since that is the distance the hoop is above the ground
so 20 = 0 + 0 + 1/2 (16) t^2
so t = sqrt(20/8)

[quick]

however by using the answer and the x position formula,
7.62 = 3t
t = 2.54
but obviously i won't be able to get that using the answer