Motion Q: Dropping/Throwing Stone in Water-Vi, Vf, a, t, d

[Jeff97]

Homework Statement

I'm asked to calculate a stone being dropped from rest to hit the water it takes 2.7s.

Now they throw the stone and it takes only 2.3s to reach the water what's was the speed they threw it at?

Relevant Equations

Vf=Vi+at d = 0.5 * g * t2

What I know for Number 1. t=2.7s d=? Vi=0m/s^-1 a=9.8m/s^-2 Vf=? Equation to use? Vf=Vi+at= 0+9.8m/s^-2x2.3= 26.46m/s So for number one the final velocity is 26.46m/s d = 0.5 * g * t2 = 0.5x9.8x2.7^2=35.721

Number 2 I know t=2.3s d=35.721 vi? Vf? a=9.8? what formula do I use?

 

[haruspex]

t=2.3s d=35.721 vi? Vf? a=9.8? what formula do I use?

In the commonest form of the SUVAT equations there are five variables, s, t, a, v_i and v_f. Five equations each omit one and use the other four.
Which of the five variables is not interesting? Which equation does that imply?

 

[Jeff97]

I was hoping you'd be able to assist me with this. But since you asked me I'll try my best. Are we finding acceleration or Vf or Vi this is what I'm unsure about? As I know s=35.7 t=2.3s I also believe I know acceleration to be $$9.8m/s^-2$$ I seem to only know 3 variables? As for the answer to this question I'm thinking it's either 4.244m/s I haven't had s similar question before so could use some help!

 

[haruspex]

Are we finding acceleration or Vf or Vi

It asks for the speed at which the stone was thrown, so v_i. So do you know a SUVAT equation involving that and s, a and t? (I know you do!)

 

[Jeff97]

$$s=ut+1/2at^2$$ This could be wrong but I presume it correct because it contains s,a,t and Vi(u) and doesn't contain Vf which we don't know and aren't asked to solve during this question?

 

[haruspex]

$$s=ut+1/2at^2$$ This could be wrong but I presume it correct because it contains s,a,t and Vi(u) and doesn't contain Vf which we don't know and aren't asked to solve during this question?

That's the one.

 

[neilparker62]

Alternate method using average velocity. Note that for uniform acceleration, average velocity occurs at the midpoint of the time interval separating ##v_i## and ##v_f##. ie $$v_{av}=v_i+\frac{aΔt}{2}=v_f-\frac{aΔt}{2}$$
First throw: $$v_{av1}\times2.7=d $$ Second throw: $$v_{av2}\times2.3=d$$ Hence: $$\frac{v_{av2}}{v_{av1}}=\frac{2.7}{2.3} $$ with ##v_{av1}=13.23 ms^{-1} ##

 

[Jeff97]

s=ut+1/2at^2 u=s/t - 1/2 at u=35.75/2.3-1/2x9.8x2.3= 4.27347826087 so about 4.3m s^1 ? is this the initial velocity he threw the ball at?

 

[Jeff97]

I've just found another problem corresponding somewhat to this: He then watches a leaf move down the stream at 0.30ms^1. He wants to drop a stone onto the leaf. Determine the position of the leaf at the instant when he must drop the stone.

 

[haruspex]

s=ut+1/2at^2 u=s/t - 1/2 at u=35.75/2.3-1/2x9.8x2.3= 4.27347826087 so about 4.3m s^1 ? is this the initial velocity he threw the ball at?

Yes.

 

[haruspex]

I've just found another problem corresponding somewhat to this: He then watches a leaf move down the stream at 0.30ms^1. He wants to drop a stone onto the leaf. Determine the position of the leaf at the instant when he must drop the stone.

This is rather different. Please start a new thread.

 

[Jeff97]

@neilparker62 why is your answer different from mine?

 

[haruspex]

@neilparker62 why is your answer different from mine?

@neilparker62 did not provide a finished answer. That analysis only went as far as finding the average speed in the thrown case. You need to combine that with an expression for the increase in speed in order to extract the initial speed.