Kinematics: falling stone question

[AStupidHippo]

Homework Statement

While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 6.91 m/s. The stone subsequently falls to the ground, which is 16.5 m below the point where the stone leaves your hand. At what speed does the stone impact the ground? How much time is the stone in the air? Ignore air resistance and take g = 9.81 m/s2.

I solved for impact velocity fairly easily. v=19.2 m/s

I'm having trouble with finding time. I have to use an online HW submitter (Sapling Learning), but it keeps saying I'm wrong.

Homework Equations

v^2 = u^2 +2as
v=u+at

The Attempt at a Solution

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19.2 = 6.91 + 9.81(t)
12.29=9.81(t)
t=1.25 seconds
Apparently this is wrong: http://prntscr.com/cdptl8

 

[Hans Herland]

Greetings.

Could you show how you come up with t = 1.25s?

I tried the problem myself, and I do get a different answer, using your formula: v = v₀ + at
You have to make the formula be based on t instead, which gives you: t = (v - v₀) / a

Now I did the calculation twice: once for figuring out the air-time the stone uses to the top of the upward throw, and one for calculating the amount of time the stone takes from the top of the throw all the way to the bottom.

The first calculation has v₀ = 6.91
The second calculation has v₀ = 0

In the first calculation, a will be negative (assuming we've set a positive direction upwards)
In the second calculation, a will be positive.

This gives you two numbers you should be able to add together to see the total amount of air time. There's probably a better method of doing this in one calculation.

 

[AStupidHippo]

Greetings.

Could you show how you come up with t = 1.25s?

I tried the problem myself, and I do get a different answer, using your formula: v = v₀ + at
You have to make the formula be based on t instead, which gives you: t = (v - v₀) / a

Now I did the calculation twice: once for figuring out the air-time the stone uses to the top of the upward throw, and one for calculating the amount of time the stone takes from the top of the throw all the way to the bottom.

The first calculation has v₀ = 6.91
The second calculation has v₀ = 0

In the first calculation, a will be negative (assuming we've set a positive direction upwards)
In the second calculation, a will be positive.

This gives you two numbers you should be able to add together to see the total amount of air time. There's probably a better method of doing this in one calculation.

Hi!

So this is what I did: (Impact velocity - initial velocity)/a = t
(19.2-6.91)/9.81=t
This gave me t=1.25

 

[Hans Herland]

Hi!

So this is what I did: (Impact velocity - initial velocity)/a = t
(19.2-6.91)/9.81=t
This gave me t=1.25

This should give you the time it would take for the stone to drop if it was thrown directly towards the ground. Since the stone is thrown upwards, you'll need to find the time it takes for the stone to travel with initial velocity of 6.91, to impact velocity of 0 (where the stone starts dropping).

 

[AStupidHippo]

This should give you the time it would take for the stone to drop if it was thrown directly towards the ground. Since the stone is thrown upwards, you'll need to find the time it takes for the stone to travel with initial velocity of 6.91, to impact velocity of 0 (where the stone starts dropping).

Ok. So I should make v= 0 and solve for time with this: 0=6.91-9.81t
t=.7
And then I add it it to 1.25? So the answer is 1.25+.7= 1.95

 

[Hans Herland]

Ok. So I should make v= 0 and solve for time with this: 0=6.91-9.81t
t=.7
And then I add it it to 1.25? So the answer is 1.25+.7= 1.95

You should re-check the 1.25 number, since that's still based on the time it would have taken for the stone to drop at the height it was thrown in. Remember that it now drops from a higher altitude, making the drop longer than before, and the initial velocity of the drop is 0.

Since the stone was thrown up into the air, it accelerated downwards due to gravity, making it slow down and eventually "stop". When this happens, the stones velocity starts accelerating again, but this time the velocity also points downwards.

I got 0.7 seconds as well when calculating the time it takes for the stone to reach the top of the throw.

 

[AStupidHippo]

You should re-check the 1.25 number, since that's still based on the time it would have taken for the stone to drop at the height it was thrown in. Remember that it now drops from a higher altitude, making the drop longer than before, and the initial velocity of the drop is 0.

Since the stone was thrown up into the air, it accelerated downwards due to gravity, making it slow down and eventually "stop". When this happens, the stones velocity starts accelerating again, but this time the velocity also points downwards.

I got 0.7 seconds as well when calculating the time it takes for the stone to reach the top of the throw.

Yes. It's 1.25+.7+.7=2.6. This is the correct answer. Thank you!

 

[LogarithmLuke]

Alternatively you could use the v=vo +at formula, but have v0 be negative, because at the bottom you are moving the oppositve way of the intitial throw. If you have movement in two different different directions one of the values has to be negative.

 

[haruspex]

Alternatively you could use the v=vo +at formula, but have v0 be negative, because at the bottom you are moving the oppositve way of the intitial throw. If you have movement in two different different directions one of the values has to be negative.

Indeed, this is the correct way to solve the problem. Since acceleration is constant throughout, it is quite unnecessary to break it into separate upward and downward stages.

 

[needhelppls]

how did you find the impact velocity?

 

[haruspex]

how did you find the impact velocity?

See the relevant equations in post #1.

 

[neilparker62]

I think OP's original solution was fine - just need to be careful not to round off answers too early.

 

[haruspex]

I think OP's original solution was fine - just need to be careful not to round off answers too early.

As in post #1? The correct answer is given in post #7.

 

[neilparker62]

As in post #1? The correct answer is given in post #7.

Apologies. Had OP used -6.91 as initial velocity instead of 6.91, the approach would have been correct. Although he should probably use the non-rounded answer to the first part (impact velocity). Or at least maintain all calculations correct to 2 dp consistent with g=9.81 m/s^2.

As you've said there was no real need to break the time problem into two parts but many students seem to prefer doing things that way.

One could also use $v_{av}Δt=16.5$ where $v_{av}=\frac{-6.91+19.27}{2}$