Velocity Related Problem -- A body is thrown upward....

[Shahab Mirza]

Homework Statement

A body is thrown upward and remains in air for 10 seconds , the initial velocity at which it was thrown in is?

The Attempt at a Solution

g = -10 , t = 10 s , vf = 0 , vi = ?

Vf = vi + gt
0=vi + (-10) (10)
Vi = 100

but its answer is not 100 its answer is 50 help me please , where I did a mistake ? Thanks[/B]

 

[Physics Newbie]

Here u got to consider the motion in two parts,
1. From down to up and
2. Up to down

From down to up we have u=u, v=0 (velocity at top is 0, it is momentarily at rest at that point), (Here)t=10/2=5s, g=-10m/s(as direction of u is opposite to acc or force of gravitation) so we have:
v=u+at
0=u-10*5
u=50

Also when you will check motion from up to down you will get the same 'u', it signifies that in a projectile motion velocity at two points lying on the same horizontal line in up and down motion will have the same velocity.

Or use formula of Projectile motion T=2usinx/t, here x=90 degree so 10=2u/10 or u=50m/s

 

[Shahab Mirza]

Here u got to consider the motion in two parts,
1. From down to up and
2. Up to down

From down to up we have u=u, v=0 (velocity at top is 0, it is momentarily at rest at that point), (Here)t=10/2=5s, g=-10m/s(as direction of u is opposite to acc or force of gravitation) so we have:
v=u+at
0=u-10*5
u=50

Also when you will check motion from up to down you will get the same 'u', it signifies that in a projectile motion velocity at two points lying on the same horizontal line in up and down motion will have the same velocity.

Or use formula of Projectile motion T=2usinx/t, here x=90 degree so 10=2u/10 or u=50m/s

Thanks a lot sir .

Oh so it means that T =10 seconds was the time ball was in air it means that it was time taken from throwing ball upward and then catching it in hand hence where the ball momentarily stops , the time till then is 5 seconds .

It means I didn't understood question correctly , the equation I used was fine , Thanks alot

 

[Orodruin]

Here u got to consider the motion in two parts,

This is not true, it is perfectly fine to consider the motion as one single parabola.

g = -10 , t = 10 s , vf = 0 , vi = ?

Where did you take vf = 0 from? This would imply that the body hits the floor at zero velocity.

 

[Physics Newbie]

This is not true, it is perfectly fine to consider the motion as one single parabola.
Where did you take vf = 0 from? This would imply that the body hits the floor at zero velocity.

vf is the velocity at the max height of the parabolic path bro.

 

[Shahab Mirza]

This is not true, it is perfectly fine to consider the motion as one single parabola.
Where did you take vf = 0 from? This would imply that the body hits the floor at zero velocity.

Vf is velocity at max height where ball stops for a while

 

[Orodruin]

vf is the velocity at the max height of the parabolic path bro.

It does not need to be and it was not clear in the OP. The natural thing to do in this problem would be to consider the full parabolic motion from the initial time until the time the ball hits the ground with the same velocity (but with opposite sign) as the initial one. Your assertion that you need to consider it as two separate motions is simply not true.

Vf is velocity at max height where ball stops for a while

The ball never stops "for a while". It will only be at rest momentarily. If you want to do it this way, then you need to consider that it will take half the time to reach the top as already pointed out. The alternative is to consider that on the way down, the body will hit the ground with the same speed as it was thrown, but with the velocity directed in the opposite direction, i.e., vf = -vi. This will take the full 10 seconds.

 

[Shahab Mirza]

It does not need to be and it was not clear in the OP. The natural thing to do in this problem would be to consider the full parabolic motion from the initial time until the time the ball hits the ground with the same velocity (but with opposite sign) as the initial one. Your assertion that you need to consider it as two separate motions is simply not true.The ball never stops "for a while". It will only be at rest momentarily. If you want to do it this way, then you need to consider that it will take half the time to reach the top as already pointed out. The alternative is to consider that on the way down, the body will hit the ground with the same speed as it was thrown, but with the velocity directed in the opposite direction, i.e., vf = -vi. This will take the full 10 seconds.

It means that I should apply time equation of projectile . It will be correct option ?

 

[Physics Newbie]

Here u got to consider the motion in two parts,
1. From down to up and
2. Up to down

From down to up we have u=u, v=0 (velocity at top is 0, it is momentarily at rest at that point), (Here)t=10/2=5s, g=-10m/s(as direction of u is opposite to acc or force of gravitation) so we have:
v=u+at
0=u-10*5
u=50

Also when you will check motion from up to down you will get the same 'u', it signifies that in a projectile motion velocity at two points lying on the same horizontal line in up and down motion will have the same velocity.

Or use formula of Projectile motion T=2usinx/t, here x=90 degree so 10=2u/10 or u=50m/s

Okay let me correct it a bit...Projectile is motion in 2D, you can divide this motion along separate x and y axis, and apply equations of constant acc. accordingly along both directions.
If you have any problem getting it, go through the derivation of projectile formulas.

 

[Physics Newbie]

From down to up we have u=u, v=0 (velocity at top is 0, it is momentarily at rest at that point), (Here)t=10/2=5s, g=-10m/s(as direction of u is opposite to acc or force of gravitation) so we have:
v=u+at
0=u-10*5
u=50

Also when you will check motion from up to down you will get the same 'u', it signifies that in a projectile motion velocity at two points lying on the same horizontal line in up and down motion will have the same velocity.

Or use formula of Projectile motion T=2usinx/t, here x=90 degree so 10=2u/10 or u=50m/s

 

[Physics Newbie]

Okay let me correct it a bit...Projectile is motion in 2D, you can divide this motion along separate x and y axis, and apply equations of constant acc. accordingly along both directions.
If you have any problem getting it, go through the derivation of projectile formulas.

So either you take a single parabolic path or take motion along x and y axis, its one and the same thing.

 

[Shahab Mirza]

So either you take a single parabolic path or take motion along x and y axis, its one and the same thing.

Yes thanks a lot