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

• Shahab Mirza
In summary: Then you can apply time equation of projectile for finding initial velocity of the motion in each part of the motion.
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]

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
Physics Newbie said:
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

Physics Newbie said:
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.

Shahab Mirza said:
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.

Orodruin said:
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.

Orodruin said:
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

Physics Newbie said:
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.

Shahab Mirza said:
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
Orodruin said:
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 said:
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.

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 said:
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.

Physics Newbie said:
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

## What is the equation for velocity of a body thrown upward?

The equation for velocity of a body thrown upward is v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity, and t is the time.

## How does air resistance affect the velocity of a body thrown upward?

Air resistance or drag can decrease the velocity of a body thrown upward by opposing its motion, causing it to slow down more quickly. This can be taken into account by using the equation v = u + at - 1/2gt^2, where g is the acceleration due to gravity and t is the time.

## What is the maximum height reached by a body thrown upward?

The maximum height reached by a body thrown upward can be found using the equation h = ut - 1/2gt^2, where h is the maximum height, u is the initial velocity, g is the acceleration due to gravity, and t is the time. This assumes that air resistance is negligible.

## How does increasing the initial velocity affect the time of flight of a body thrown upward?

Increasing the initial velocity of a body thrown upward will increase the time of flight, as the body will have a higher velocity and will take longer to slow down and return to the ground. The relationship can be represented by the equation t = 2u/g, where t is the time of flight, u is the initial velocity, and g is the acceleration due to gravity.

## What is the acceleration due to gravity on Earth?

The acceleration due to gravity on Earth is approximately 9.8 m/s^2. This value can vary slightly depending on location and altitude, but is generally considered to be a constant for most calculations involving free fall and projectile motion.

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