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.