Free Fall Question (Non-solvable 2nd order equation)

[phys_student1]

Hi,

This is from "3000 solved problems in physics". Yes, I have the solution but
I still have a question.

a ball is thrown vertically upward with a velocity of 20 m/s from a top of a tower
having height of 50m. On its return it misses the tower and finally hits the ground.

What time t elapses from the moment the ball is thrown until it passes the edge of the tower?

The book solution is clear, it chooses y=0 to be at the top of the tower.
Then: y=v_ot + (1/2)at²
This is solved to give t=0 (the beginning) or t=4.08s.

OK, nice, but I'd like the y-axis to be at the ground. Clearly, I cannot apply the above equation,
because it will neglect the additional potential energy that the ball have when now it's 50m above
the y axis. (If I substituted 50 for y in the eqn, I'll get a non-solvable eqn, or, if you like, an imaginary solution).

So how can we solve this (without using the concept of energy) ?

 

[HallsofIvy]

Taking y= 0 at the bottom of the tower means each point will have "y" 50 higher than before- that will give you $y= v_0t+ (1/2)at^2+ 50$. And, since the problem asks for the time the ball passes the tower again, you want to know when the ball will again have height 50. That is, you want to solve
$$y= v_0t+ (1/2)at^2+ 50= 50$$

Subtract 50 from each side and you have exactly the same equation as before.

 

[Doc Al]

Just to add to what Halls' explained:

Then: y=v_ot + (1/2)at²

This equation assumes that y = 0 at t = 0. A more general equation would have an arbitrary initial position y₀:
y = y₀ + v₀t + (1/2)at²

 

[phys_student1]

Thanks a lot !

Yes, I should have used (delta y), rather than (y).

Many thanks again.

 

[rwisz]

Hello,

Thank you for your question. I understand your concern about wanting to have the y-axis at the ground level instead of at the top of the tower. In this case, we can still use the equation y=vot + (1/2)at^2 to solve for the time t, but we just need to adjust our initial position and final position values.

Since we want the y-axis to be at the ground level, we can set y=0 to be at the ground level instead of at the top of the tower. This means that our initial position (yo) will now be -50m (negative because it is below the y-axis) and our final position (y) will be 0m. So our equation becomes:

0 = -50 + 20t + (1/2)(-9.8)t^2

Solving for t, we get t= 4.08s or t = 0s. However, we know that t=0s is not the correct answer since the ball is thrown at the beginning of the experiment. So the correct answer is t=4.08s, which is the same as the book solution.

In this case, we did not have to use the concept of energy to solve the problem. We simply adjusted our initial and final position values to be consistent with our chosen y-axis.

I hope this helps clarify your question. If you have any further questions, please let me know.

Best,