Question

I need a quick help here, it's about acceleration of a falling object, ignoring air resistance;

I have to calculate a baseball's falling speed, just right before it hits the ground, I did everything and found that this ball with start falling from 48.42 meters. (Exactly at \(\frac{4745}{98}\)

So from what I understood from class and book, this object has no more speed when it starts to fall, therefore, no more kinetic energy and has maximum potential energy.

I thought of find the speed with the formula \(\large E_k = \frac{1}{2}mv^2\) but I'm really not sure about it;

Answer 1

So I tried to do it with the Earth's acceleration, which is \(\large 9.8\frac{m}{s^2}\)

Answer 2

but I'm don't know how, can someone guide me through this please? Thanks :)

Answer 3

@KingGeorge
@cshalvey

Can you help me please? :D

Answer 4

Have you tried using \(-9.8 m/s^2\)?

Answer 5

Nope, I don't know how it works D:

Answer 6

Well, if you're asked for speed, you should use the absolute value of velocity, and your way appears to be valid. However, if you're asked about velocity, you should use \[-9.8\frac{m}{s^2}\]as your acceleration instead.

Answer 7

Let me see what I get when I use a negative acceleration.

Answer 8

I'm getting a speed of \(\approx30.8064\) and a velocity of about -30.8064

Answer 9

I got that as well, with \(\large E_k = \frac{1}{2}mv^2\)

Answer 10

but can you show me how to got 30.8064 with the acceleration factor? :D

Answer 11

Use the formula \[v_f^2=v_0^2+2ad\]Where \(v_0\) is the initial velocity, \(a\) is acceleration, and \(d\) is the distance traveled.

Answer 12

I've got to go to dinner now, but I'll be back to answer more questions later

Answer 13

Alright, I think I've understood that acceleration thing now, thank you very much! :]

Answer 14

An easy way to calculate this is to simply use:\[PE_{top}=mgh\]\[KE_{ground}=\frac{1}{2}mv^2\]Conservation of energy states that PE will equal KE in this situation so:\[mgh=\frac{1}{2}mv^2\]Simplifying you get: \[\not mgh=\frac{1}{2}\not mv^2\]\[(9.8m/s^2)(48.42m)=\frac{1}{2}v^2\]Now just solve for v:\[v=\sqrt{(2)(9.8m/s^2)(48.42m)}\]\[v=30.8m/s\]

Answer 15

Note at the top that I meant \[\Delta KE~will~equal~ \Delta PE\]

Answer 16

Never thought of conservation. It's all clear now, Thanks a lot! :D

Answer 17

Cool. Also note that if you consider downward negative in this problem the final value will obviously be -30.8 m/s.

Answer 18

Yes, I got that part, thanks to KingGeorge, thank you both! :)