Question

A 0.145 kg baseball is thrown from a platform above the ground at a 10 m/s. The ball is moving 15 m/s when it strikes the ground. From what height above the ground was the ball thrown?

Answer 1

@Loser66 Here is the new question :)

Answer 2

find t from v = v_0 + at where a = g = 9.8 m/s^2
then plug t into the formula h = h_0 + v0t + 1/2 at^2 where h0 is the height we need to find and h =0 ( when hitting the ground, the ball is get 0 height)

Answer 3

So the first part is 10 * 9.8 * t = 15

Answer 4

So t = 0.15?

Answer 5

Is clear what the initial direction of the ball was? We may need vector addition here.

Answer 6

I don't think so... But I guess it's possible...

Answer 7

@wolfe8 , could you help me finish? Sorry I keep asking for your help :P Hope I'm not a bother.

Answer 8

That is supposed to be an addition not multiplication 10 + 9.8 * t = 15

Answer 9

Okay.. So it would really be -4?

Answer 10

Actually -5 makes more sense :P Sorry

Answer 11

I guess we assume it is thrown horizontally.

No you cannot have negative time :P See, first you subtract both sides with 10 then divide with 9.8

Answer 12

OH DUH! Silly me! I've been doing to much math today haha My brain is fried!

Answer 13

0.51 = t :P

Answer 14

\[0 = h _{i} + 10 * 0.51 + \frac{ 1 }{ 2 } * 9.8 * 0.51^2\]

Answer 15

Does that look okay?

Answer 16

\[h _{i} = -6.4?\]

Answer 17

Yup it will be the magnitude so 6.4 m. We get negative because we took acceleration downward (g) as positive in the calculations. Am I right, @Loser66 ?

Answer 18

Loser66 signed off earlier, but I'm glad you agree! So my answer would be about 6 m?

Answer 19

Wait, or maybe we have to use -g in that.

Answer 20

Hm... That would give an answer of about 4, correct?

Answer 21

@wolfe8 - If the initial velocity is horizontal then it does not affect the vertical velocity.

As Douglas says - there may be a vector question here.

In the absence of info I personally would have assumed that the initial velocity was vertically downwards.
v = u+at
solve for t

then
s = ut +0.5at^2

u and g and s have the same direction so solve for s.

s is the distance below the point thrown.

Answer 22

HMMMMMMM. I am getting confused. Well since we used positive for both equations, I think we just go with positive 6.4 m.

Yea that's what I was thinking but I decided to just work with what Loser66 said. @MrNood

Answer 23

Huh, maybe @MrNood is right

Answer 24

Which is which variable, again? I'm getting confused with all of these numbers and letters O.o g is gravity... S is the distance below the point thrown... what is u?

Answer 25

v= final velocity
u = initial velocity
s = displacement (distance)
t=time

Answer 26

So t is still 0.51

Answer 27

yes

Answer 28

Yea they're the same equations and what Loser said is the same thing. I just didn't realize it.

Answer 29

So my answer is about 6.3

Answer 30

Okay, so it is still the same!
Good! I was scared for a minute there

Answer 31

Thank you for your help everyone!!!

Answer 32

You're welcome. I just helped with the math btw. I went with Loser's physics. And MrNood confirmed it.

Answer 33

You were all a big help :P! I needed all three of you! haha :)