Can You Calculate the Time Needed for a Model Rocket to Pass Through a Hoop?

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To calculate the time needed for a model rocket to pass through a hoop 20 m high, the rocket's vertical motion must be analyzed using the equation for displacement under constant acceleration. The thrust of 8 N results in an acceleration of 16 m/s² when divided by the rocket's mass of 0.5 kg. The correct equation for vertical displacement is 20 = 0 + 0 + 1/2(16)t², leading to t = sqrt(20/8), which gives approximately 1.58 seconds. However, when calculating the horizontal distance using x = 3.0t, the time derived from the vertical motion does not yield the expected horizontal distance of 7.62 m. The confusion arises from the need to reconcile both the vertical and horizontal components of the rocket's motion to determine the correct timing and positioning for passing through the hoop.
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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?
 
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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?
 
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?
 
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thats exactly what i want to find out. but i already know the distance, i need to find how to get that answer.
 
(read above post)
 
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)
 
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
 
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