A rocket is fired up and ascends w/ a constant vertical acceleration 4.0m/s^2 for 6.0s.FindVelocityWithWhichTheRocketHitGround

Question

Physics question 

Answer 1

For this problem, we need to break it up into 2 parts - while the rocket is accelerating, and then afterwards.

 

While the rocket is accelerating

 

We are given the time and acceleration. The initial velocity is implied to be 0 (since all rockets start from rest). We will end up needing the final velocity and the final height to work on the next part. Let's choose a coordinate system so that y=0 on the ground and UP is the positive direction (we will keep this coordinate system for the next part, too).

 

First, we find the height during this phase by using the position function.

 

y(t) = yo + Vo*t + (1/2) * a * t^2

 

We want y(t=6), using Vo = 0 and a = 4.0, which looks like...

 

 

y(6s) = 0 + 0 *(6) + (1/2) * 4 * 6^2
         = 0 + 0 + 72
y(6s) = 72 m

 

Now, we want the velocity at the end of those 6 seconds.

 

v(t) = Vo + a * t

 

V(6) = 0 + 4 * (6)

       = 24 m/s

 

Now that we have those numbers, we can "reset" our variables moving on to the next stage.

 

Rocket is no longer accelerating (from its engines)

 

What is the acceleration of an object that is flying through the air? That's a similar question to "what is the reason objects fall back to the ground?"

 

Gravity, of course. The value of the acceleration due to gravity is 9.8 m/s^2. But let's ask ourselves, what direction does it point? Down. Remember when we said we were going to use the same coordinate system for the whole problem? "Up" was positive then, so "down" has to be negative.

 

a = - 9.8 m/s^2

 

For our "initial" variables, we use what we found as the "final" ones from the last part.

 

Vo = 24 m/s

Yo = 72 m

 

We are asked to find the velocity as the rocket crashes back into the ground. So what is Δy (the change in height)? If we start at 72 m and end up back on the ground at 0m, 

 

Δy = - 72 m

 

Now, we have an initial velocity, acceleration, a change in position, and we want to find a final velocity. Which kinematic can help us with that? Note the we don't know the time, so we pick the equation that doesn't have time in it.

 

V^2 = Vo^2 + 2 * a * Δy

 

Plugging in everything we know looks like...

 

V^2 = 24^2 + 2 * (-9.8) * (-72)

 

V^2 = 1987.2

 

V = 44.58 m/s

 

We are almost finished now. We defined the positive direction to be up, but what is the direction of the velocity as it crashes into the ground? Down. So, the answer is 

 

Vf = - 44.58 m/s

 

Round to 2 sig figs if your teacher requests that.

 

Side note: the equation we used for the last part has the velocity^2, which means that both the positive and negative values are valid, as the negative disappears upon squaring it. So -44.58 is a solution as well, but I glossed over that fact for clarity.

 

I hope this helps!