# One dimensional motion - Why does a ball fall back in my hands when I throw it up in a moving bus? [duplicate]

Suppose you are moving in a bus and you throw a ball upwards. Why does it always land in your hand and does not fall behind you? The bus is moving with a uniform velocity in a straight direction.

Now if we consider that the bus is constantly accelerating. Will the ball fall back again in my hands? I think that the answer is no. Why so? Why doesn't the ball retain its acceleration when it retains its original velocity?

## marked as duplicate by John Rennie newtonian-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 2 '16 at 15:03

If you are sitting in a moving Bus say going in x-direction or any other vehicle and you toss a ball up in your moving frame at time t(1) the ball will move up no doubt but instantaneous velocity in the forward direction (velocity of the bus ) will also be there with the ball.

If the bus is moving with uniform velocity the ball sent up will be moving under gravity and fall back again after time t(2) . so naturally it will fall down in your hand; as with uniform speed you have also moved the same distance as the ball in forward x direction. yours and ball's velocity along x is same.

If however you are accelerating in the forward direction (due to acceleration of the Bus) then by the time ball has moved up and came down under gravity the distance traversed by you will be more than the ball in the forward direction. the Ball's velocity in forward direction was the same when it had left your hand to go up.

However your motion in forward direction is accelerated so you will move ahead each second with added velocity and the ball will fall back behind you.

The ball can not have accelerated motion in forward direction as no forces are acting in the horizontal forward direction so no momentum change can take place. To retain accelerated motion one needs action of force and its work done on the body leads to change in the velocity i.e. its momentum and K.E..

The ball will come towards you while the bus accelerates, because being in the air, it is not linked to anything, therefore, the bus cannot transmit its growing kinetic energy to it. Therefore, the ball will tend to oppose to the bus' movement and will seem to be going backwards, relative to a bus passenger's point of view. So it's all about point of views, which is the common term for reference frame :

When in uniform linear motion relative to an inertial reference frame, the bus is itself an inertial reference frame. Therefore, if you throw a ball with an vertical initial velocity, it will go up to a maximum altitude and fall in free fall like it was "on earth". Following newton's second law of motion : $$\sum{\vec{F_i}} = m.\vec{a}$$ With $m$ being the ball's mass, $\vec{a}$ its acceleration relative to the bus reference and $\sum{\vec{F_i}}$ the sum of all forces applied to the ball, which in case of a free fall is reduced to the gravitational force only : $\sum{\vec{F_i}} = m\vec{g}$

Now, if the bus is now accelerating (relative to the earth's inertial reference frame), it becomes a non-inertial reference frame, which is way more fun because now, the second law of motion has inertial parameters : $$m.\vec{a} = \left(\sum{\vec{F_i}}\right) + \vec{F_e} + \vec{F_c}$$ With $\vec{F_c} = -m.\vec{a_c}$ being the Coriolis effect, which in case of an accelerating bus is non-existent.

However, $\vec{F_e} = m.\vec{a_e}$ is more important here, because that's the mathematical model of "why is the ball likely to hit me in the face while the bus is accelerating". $F_e$ is called a fictious force, because it's a force that does not come from any physical phenomenon, but is used to make the second motion law model working in non-inertial frames, compensating for the non-inertial frame acceleration. It is equal to the object's mass ($m$) times the non-inertial frame's acceleration relative to an inertial frame. Let's say the bus has a constant acceleration of $\vec{a_b}$, the second law of motion applied to the ball in free fall is now : $$m.\vec{a} = m.\vec{g} - m.\vec{a_b}$$ Or : $$\vec{a} = \vec{g} - \vec{a_b}$$ Which translates in : while the ball is accelerated by gravitation, it also accelerates in the same amount of the bus, but in the opposition direction.