What is the initial speed of the rock when thrown off a cliff?

[Let It Be]

1. When a kid drops a rock off the edge of a cliff, it take s 4.0s to reach the ground below. When he throws the rock down, it strikes the ground in 3s. What inital speed did he give the rock?


ΔX=ViTf + 1/2ATf^2
H=1/2gt^2
Vf^2=Vi^2+ 2AΔX

3. Hi everyone! I know that when the kid drops the rock, you can use the free fall equation- H=1/2gt^2; but the problem doesn't say the height of the cliff and I don't know "g", unless it's just 9.8? Also, I don't know inital velocity, so I just don't know how to go about this! If anyone can help me, that'd be much appreciated!

 

[SammyS]

Assuming the cliff is on the earth, then yes, g = 9.8 m/s².


When it says the kid drops the rock, you can assume that the initial velocity is zero.

 

[Let It Be]

Hmm...why would there be two different scenarios though? One with the kid dropping the rock, and the other him throwing it? And I don't understand what the question is asking with "intial speed", is that just inital velocity? Then wouldn't they both be zero?

 

[gneill]

Hmm...why would there be two different scenarios though? One with the kid dropping the rock, and the other him throwing it?

Because that is part of the problem that is being posed.

And I don't understand what the question is asking with "intial speed", is that just inital velocity? Then wouldn't they both be zero?

When you throw something it's initial speed isn't zero. It's whatever speed it leaves your hand.

Also, velocity is just a fancy speed: it has magnitude and direction In this case you can take the initial speed to be directed downwards and call it the initial velocity if you like.

I suggest that you use the first scenario (rock is dropped) to find the height of the cliff. Then use the first of your equations to wrangle with the second scenario.

 

[Let It Be]

Good advice!

Does this look right?

H=1/2gt^2
H=1/2(9.8)4s
H=19.6m

ΔX=ViTf + 1/2 ATf^2
19.6=Vi (4) + 1/2(o)Tf^2
Vi=4.9m/s

Thanks

 

[gneill]

Good advice!

Does this look right?

H=1/2gt^2
H=1/2(9.8)4s
H=19.6m

Whoops. You forgot to square the time; that's a t² up there.

 

[Let It Be]

Whoops. You forgot to square the time; that's a t² up there.

Ahh, I gotcha. Were those the right steps to solve it though?

 

[Let It Be]

Wait...does the 3s from when he throws the rock ever come into play?

 

[gneill]

Wait...does the 3s from when he throws the rock ever come into play?

Of course. It takes 3s to hit bottom when he throws it rather than drops it. That's the effect of the initial velocity.

 

[Let It Be]

Of course. It takes 3s to hit bottom when he throws it rather than drops it. That's the effect of the initial velocity.

So in the ΔX=ViTf+1/2ATf2 the Tf would be 3s not 4s?

 

[gneill]

So in the ΔX=ViTf+1/2ATf2 the Tf would be 3s not 4s?

Do you think it would take the same time for both scenarios?

 

[Let It Be]

Do you think it would take the same time for both scenarios?

No..so do this problem twice to find his inital speed for both scenarios, using 4s for the first and 3s for the second?

 

[gneill]

No..so do this problem twice to find his inital speed for both scenarios, using 4s for the first and 3s for the second?

When you DROP an object its initial speed is zero. The initial scenario allows you to calculate the height of the cliff. Use this height in the second scenario.