Suppose a ball is thrown vertically upwards. Will it take the same time to go up and come down again? Why?

Answer 1

No it not take the same time, because when you through the ball with velocity V suppose, then in the way it has to opposse the air and the velocity will be decreases, So it will take more time to come up and down.
The time will depend on the air and the size and mause.
as much the all things it will take more time.

Answer 2

The answer is A When you through the ball up in the air it has a velocity/speed of X. Gravity will slow the ball until it has no more upward velocity/speed. Gravity will start pulling the ball back down again, increasing in velocity/speed until you catch it again. Gravity is a constant. The distance between your hands and the point at which the ball stopped is the same for both the upward and downward trip. It takes the same amount of time to slow the ball as it does to speed it back up while falling. SR

Answer 3

EDIT: THIS ANSWER HAS BEEN CORRECTED. Please note the change. Additional edit 12/22.

It is true that in the absence of air friction the time to go up is the same as the time to go down. However, if air friction is taken into account, the situation gets very complex: the forces on the ball are not symmetrical. On the way up, gravity and drag both act in the same direction to slow down the ball. On the way down, gravity and air friction are in opposite directions. It is not immediately obvious which time will be longer, but in general they are not the same. The time to go down is always greater that the ascent time, the amount depending on the launch velocity. The difference isn't much, about 10% more if launch velocity is 90% of terminal velocity.

If you are interested in the derivation of this result, you can try to slog through the following four pages of math (the key result is on the last page):

http://img126.imageshack.us/img126/4903/trajectoryupdown1vw8.png
http://img156.imageshack.us/img156/9081/trajectoryupdown2fw3.png
http://img184.imageshack.us/img184/3717/trajectoryupdown3vf1.png
http://img250.imageshack.us/img250/8544/trajectoryupdown4uw2.png

I did this because your question roused my curiosity and I thought the answer was not simple.

Answer 4

Yes it will. This is because it is acting under *constant deceleration*. The only force acting on it once it has been given initial velocity (i.e. you throwing it) will be gravitational,which pulls it down with a constant value of 9.8 ms^-2. This will produce a parabolic velocity/time graph, which is symmetrical and from that we can deduce that the time before the peak will be the same as the time after the peak.

Answer 5

No.

The ball can be launched at any number of speeds, but due to the gravitational constant the ball will fall at a velocity determined by its altitude when the fall begins. This is the case with a tru vertical trajectory.

However, if the trajectory arcs sufficiently, the ball can exceed terminal velocity.

Answer 6

The same force which ( decelerates) stops the ball at certain height pulls it down ( accelerates) through out this height during the return path.

Therefore the time will be the same.

Answer 7

Yes the time taken by the ball will be same. if you apply the famous equation of motion S=ut+1/2gt2, where the symbols have their usual meanings you will find that the time taken is same as the distance travelled is same.

Answer 8

When you throw something upwards, you are acting "against" gravity;
when something comes down, it's "encourages" gravity.
Hence for the upward travel will take more time than when comes down.

Correct me if i'm wrong

Answer 9

If you disregard air resistance, yes. Otherwise no. If air resistance is taken into account, it will take longer to come down than to go up.

Answer 10

On the third try gp4rts finally gets it right.