One dimensional motion- object accelerating straight downwards

[texan14]

One dimensional motion-- object accelerating straight downwards

Homework Statement

A rocket, initially at rest, is fired at "t = 0" vertically down from a building of height "H". The rocket's acceleration, including the effects of gravity, is downwards with increasing magnitude given by a(t) = βt, where "β" is a known constant. When does it hit the ground and how fast is it going when it hits?

Homework Equations

x_f = x_i + v_i*t + (1/2)*a*t²

The Attempt at a Solution

x_f = (0) + v_i*t + (1/2)*β*t²

I plugged everything into this equation, but it doesn't look right. I'm really confused. thanks in advance

 

[ehild]

Hi Texan14,

Is the acceleration constant? If not, why do you want to use the formula for constant acceleration?

Go back to the definition of acceleration.

ehild

 

[texan14]

I decided to integrate until I got the position function and solved for "t" and just plugged that into v(t). Thank you for your help, I don't know why I wanted to use that equation! haha

 

[ehild]

It does not hurt to read the the problem before plugging in everything to everywhere...

ehild

 

[asteriatic]

I would approach this problem by first identifying the known variables and the given information. From the problem statement, we know that:
- The initial position of the rocket (xi) is at the top of the building, which we can assume to be the origin (xi = 0).
- The final position of the rocket (xf) is at the ground, which we can assume to be a height of 0.
- The initial velocity of the rocket (vi) is 0, since it is initially at rest.
- The acceleration of the rocket (a) is given by a(t) = βt, where t is the time and β is a known constant.

Using the equation xf = xi + vi*t + (1/2)*a*t^2, we can substitute in the known values and solve for t:
0 = 0 + 0*t + (1/2)*β*t^2
0 = (1/2)*β*t^2
0 = t^2

This equation does not have a single, unique solution for t. However, we can interpret this result as meaning that the rocket will hit the ground at two different times: t = 0 and t = ∞ (since the rocket will continue to accelerate downwards indefinitely).

To find the speed of the rocket when it hits the ground, we can use the equation v = vi + a*t. Since vi = 0, we can simplify this to v = a*t. Plugging in the value of a(t) = βt, we get v = βt. At t = 0, the rocket has a speed of 0, but as t increases, the speed of the rocket will also increase. We can see that as t approaches ∞, the speed of the rocket will also approach ∞, meaning it will be going very fast when it hits the ground.

In conclusion, the rocket will hit the ground at two different times (t = 0 and t = ∞) and its speed when it hits will approach infinity. This is due to the constant acceleration downwards and the fact that the rocket is initially at rest.