Physics question about velocity, displacement, acceleration, time...?

Question

I know what the answer is, but I don't know how to get there!

Question: A ball is thrown upward from the ground w/ an initial speed of 25 m/s; at the same instant, another ball is dropped from a building 15 m high. After how long will the balls be at the same height?

Answer: 0.60 s

Doug: I agree w/ your solution. I know you start with d = vt + (1/2)at^2. I was wondering how you came up w/ the 2nd equation. Thanks.

We were never given this equation in class [h = ho + vo t + (1/2) gt^2], so I guess it would've made things a lot easier!

Answer 1

h = 25*t - (1/2)*9.8*t²
h=15 - (1/2)*9.8*t²
Set them equal and solve for t.

Doug

Answer 2

If you don't mind, I will answer the question that was ment for Doug

He used this:

h = ho + vo t + (1/2) gt^2

In the first case, ho is 0. And in the second case, vo is 0

We are accepting that the first ball always goes up. But it could happen that it would reach v=0 in the meantime and begin to fall, in this case they would never get the same high

Lets calculate when the first ball will get to v = 0

v = vo - gt

0 = 25 - 9.8 t

t = 25/9.8 = 2.55 s

Let's now see at which high it would be when he gets v=0

h = vot - (1/2)gt^2
h = 25^2.55 - (1/2)*9.8*2.55^2 = 51.3 m, so this won't happen and the balls will be at the same heigh

Ana

Answer 3

The preliminary speed is larger than the needed very final speed. (3E5 m/s) > (5.4E4 m/s) using fact the acceleration is helpful (8E14 m/s²), you will on no account get that very final speed. If the acceleration were undesirable (-8E14 m/s²), then 5.4E4 = 3E5 + t ? (-8E14) t = 3E-10 s (t = 0.3 ns (nanoseconds)) the area could be d = (5.4E4 + 3E5) ? (3E-10) / 2 d = 0.54E-4 m (d = fifty 4 nm (nanometer))

Answer 4

Can you integrate? If you can, you get one equation from the other one

Ilusion