Motion of a ball thrown vertically upwards
You must have seen someone throwing a ball vertically upwards or probably you have thrown it yourself. Yes, the ball goes up for a while and then stops and starts to come down and touches the ground with a bang if you can’t catch it. Let’s discuss the phases of this traversal and motion with some formula and examples. Certainly this motion is due to the earth’s gravity on the ball.
Considering the upward movement:
You throw the ball vertically upwards with some velocity say v1, which we will consider as the initial velocity for the upward path. After certain time period say t, the ball reaches a height beyond which it can’t move upwards anymore and stops there i.e. its velocity becomes zero at that height. The height where the velocity becomes zero which is the maximum height the ball went upward, say is H. And for this upward movement, the final velocity v2 is 0 because the ball has stopped at the end of this upward traversal.
It’s pretty evident that after the upward throw, the velocity of the ball gradually decreases i.e. a negative acceleration was working on the ball. The acceleration is negative because this acceleration is directing downwards while the ball is moving upward and because of this negative acceleration the velocity of the ball is gradually decreasing. Yes. this acceleration is nothing but the acceleration due to the earth’s gravity (or gravitational pull or force) on the ball.
Using formula of motion
As this acceleration (g) is working opposite to the upward velocity we have to use a negative sign in the formula below, used for the upward movement of the ball. We know the value of g in SI is 9.8 m/second square.
Using one of the equations of motion,
v2 = v1 – gt ……………………(i)
so 0 = v1 – gt
or, t = v1/g …………………….(ii)
So the time taken by the ball to reach the maximum height is = (Initial Velocity with which the ball is thrown vertically upwards) / (acceleration due to gravity)…..(iii)
So just for example, if a ball is thrown vertically upwards with 98 m/s velocity, then to reach the maximum height it will take = 98/9.8 =10 seconds.
And the maximum height H reached is obtained from the formula:
or H= v12/2g …………(iv)
Therefore if a ball is thrown vertically upwards with 98 m/s velocity, the maximum height reached by it would be = 98 x98 /(2 x 9.8) meter = 490 meter.
Downward movement of the ball
As the ball reaches the maximum height now it starts its free fall towards the earth. The force applied on it is again the gravity and this time its having a positive acceleration i.e. its following the same direction of the acceleration.
It can be proved that the time for the downward movement or the time taken by the ball to fall from the highest point and reach the ground is same as the time required for the upward movement = v1/g.
So for a vertically thrown object the total time taken for its upward and downward movements is 2v1/g
And one important point, during the downward fall the magnitude of the velocity of the ball just before touching the ground would be same as the magnitude of the velocity with which it was thrown upwards (v1 here).