How fast is ball thrown up to come down in 2 seconds

[David Earnsure]

Homework Statement

Ignoring air drag, how fast must you toss a ball straight up for it to take 2 seconds to return to same level you threw it from.

1. 20 m/s
2. 7.5 m/s
3. 15 m/s
4. 5 m/s
5. 10 m/s

Homework Equations

vf = vi + a x t

The Attempt at a Solution

To me this problem seems unsolvable as I have no idea about the distance the ball needs to travel, all I have is total time (2 seconds) and it acceleration on the way down, as if the ball is tossed upwards its acceleration will be different to - 9 .8m/s^2 ( I think?).

I don't want to be given the answer I just want to be pushed into the right mindset of thinking.

Thanks.

 

[JeremyG]

as if the ball is tossed upwards its acceleration will be different to - 9 .8m/s^2

No, the acceleration of the ball will be the same going up as it is coming down. Both equals to 10ms^-2.
To see why, ask yourself what forces are acting on the ball during its travel and equate it to ma (Newton's 2nd Law)

There's no need for calculations in this question. Intuitively, if the acceleration on its way up and down is the same, what does that tell you about the time it takes to travel up to its maximum height and the time that it takes to come back down to its original level?

That should set you thinking on what the throwing speed should be (with an understanding that the speed of the ball at maximum height is 0m/s, obviously).

 

[David Earnsure]

Oh thanks, that's actually really simple!

If it was thrown up at 10m/s the force of gravity on the ball would reduce that speed to 0m/s after one second, once is comes to 0m/s its reached its peak and would then fall accelerating at 10m/s^2 after one second reaching a speed of 10m/s it would be back at the same height it was thrown at only taking two seconds.

 

[JeremyG]

Yup well done! :D

 

[hamza2095]

No, the acceleration of the ball will be the same going up as it is coming down. Both equals to 10ms^-2.
To see why, ask yourself what forces are acting on the ball during its travel and equate it to ma (Newton's 2nd Law)

There's no need for calculations in this question. Intuitively, if the acceleration on its way up and down is the same, what does that tell you about the time it takes to travel up to its maximum height and the time that it takes to come back down to its original level?

That should set you thinking on what the throwing speed should be (with an understanding that the speed of the ball at maximum height is 0m/s, obviously).

How would you do this question with one of the kinematics equations? I'm guessing the vf=0, d=0, a=-9.8, t=2 and you would be solving for initial velocity, correct?

 

[SteamKing]

How would you do this question with one of the kinematics equations? I'm guessing the vf=0, d=0, a=-9.8, t=2 and you would be solving for initial velocity, correct?

Except, if you throw the ball in the air and then catch it when it comes back down, vf ≠ 0. Where might the velocity of the ball be zero after it is released?

 

[hamza2095]

Except, if you throw the ball in the air and then catch it when it comes back down, vf ≠ 0. Where might the velocity of the ball be zero after it is released?

Wouldn't the velocity after you catch it be 0 because it's stopped moving? Although I do know for sure that the final velocity is 0 when it's reached its peak

 

[SteamKing]

Wouldn't the velocity after you catch it be 0 because it's stopped moving?

That's the same as saying the velocity of the ball is zero while you're holding it but before you throw it.

It's technically true, but it is information which is not useful to solving the problem.

Although I do know for sure that the final velocity is 0 when it's reached its peak

Now, you have something you can work with.

 

[hamza2095]

That's the same as saying the velocity of the ball is zero while you're holding it but before you throw it.

It's technically true, but it is information which is not useful to solving the problem.

Now, you have something you can work with.

Ohh okay, so is this correct?
Vi = Vf - a*t
= 0m/s - (-9.8m/s^2*1s )
= 9.8m/s (t = 1 because the acceleration is uniform, so, the time it takes to go to the top from the hand, is the same as the time it takes going from the top to the hand, so you divide the time given by 2)

 

[SteamKing]

Ohh okay, so is this correct?
Vi = Vf - a*t
= 0m/s - (-9.8m/s^2*1s )
= 9.8m/s (t = 1 because the acceleration is uniform, so, the time it takes to go to the top from the hand, is the same as the time it takes going from the top to the hand, so you divide the time given by 2)

That's correct.

 

[azizlwl]

Ohh okay, so is this correct?
Vi = Vf - a*t
= 0m/s - (-9.8m/s^2*1s )
= 9.8m/s (t = 1 because the acceleration is uniform, so, the time it takes to go to the top from the hand, is the same as the time it takes going from the top to the hand, so you divide the time given by 2)

dv/dt=a
Up as positive
(-v-v)/2=-9.8
v=9.8m/s