Question

A 10 kg toy rocket accelerates upward and has a force of air resistance of 5 N acting on it. The engine generates 133 N of force.
a) What is the height the rocket will reach after 3 seconds (assume the initial velocity is 0m/s)?

Answer 1

We need to find the net force acting on the rocket.

If you draw a free body diagram, which you should ALWAYS do, you'll see that 133N acts upwards and 5N acts downwards and gravity acts downwards. From Newton's Second Law, \[\sum F_Y = 133 - 5 - m \cdot g=m\cdot a\]This equation can be solved for the acceleration, a.

Now, we can say that the force is constant and the mass is constant because we are given no information to suggest otherwise. This means that acceleration must be constant. We can use our constant acceleration kinematic equations of motion to find the height after 3 seconds. \[y(t) = y_0 + v_0 \cdot t +{1 \over 2} a t^2\]From the problem statement we know that \(v_0=0\) and we can say that \(y_0 = 0\)
Simply plug in \(t=3\) and the acceleration value found in the first equation to find the height y.

Answer 2

Good explanation...better than me posting a full answer.

Answer 3

The answer my teacher gave us is 13.5 m and i am not getting it even when i do it your way.

Answer 4

If you followed eashmore's post you should have come up with 13.5m as well.

Here's how it looks completely worked out:
\[F_{thrust}=133N\]\[F_{drag}=5N\]\[F_{grav}=(10kg)(9.8m/s^2)=98N\]\[F_{net}=133N-5N-98N=30N\]\[a=\frac{30N}{10kg}=3m/s^2\]Now plug it into the kinematic equation using t=3s:\[y(t)=y_0+V_o(t)+\frac{1}{2}at^2=0m+0m/s (3s)+\frac{1}{2}(3m/s^2)(3s)^2=13.5m\]

Answer 5

I read the mass wrong, i thought it was 1000g :s Thank You so much to both of you!!!