Question

A toy rocket is launched with an initial ve- locity of 12.0 m>s in the horizontal direction from the roof of a 30.0-m-tall building. The rocket’s engine produces a horizontal acceleration of 11.60 m>s 3 2t, in the same direction as the initial velocity, but in the vertical direction the acceleration is g, down- ward. Ignore air resistance. What horizontal distance does the rocket travel before reaching the ground?

Answer 1

D=90.3m.

Step-by-step explanation:

Here we got a parabolic motion problem, On the Y axis we have a free-fall motion, so the time to reach the ground is given by:

`y=yo+Vo*t+(1)/(2)*a*t^2\\0=30m+0*t-(1)/(2)*(-9.8)*t^2\\t=2.5s`

Having that time we now can calculate the horizontal distance traveled by the rocket.

The acceleration is a continous function, if we integrate the acceleration we obtain the velocity:

`vf=\int\limits^(t)_0 {11.60*2t} \, dt\\ vf=11.6m/s^3*t^2`

`Vx(t)=12.0m/s+11.6m/s^3*t^2`

the displacement is given by:

`D=\int\limits^t_0 {Vx} \, dx`

`D=12.0m/s*t+3.86m/s^3*t^3\\`

Substituting the time:

D=90.3 m